الرياضيــــــــــــات للصف الرابع العلمي

‫اق‬‫ﺮ‬‫اﻟﻌ‬ ‫ﺟﻤﻬﻮرﻳﺔ‬
‫اﻟﺘﺮﺑﻴﺔ‬ ‫ارة‬‫ز‬‫و‬
‫ﻟﻠﻤﻨﺎﻫﺞ‬ ‫اﻟﻌﺎﻣﺔ‬ ‫اﻟﻤﺪﻳﺮﻳﺔ‬
‫اﻟﻌﻠﻤــــــﻲ‬ ‫اﺑــــــﻊ‬‫ﺮ‬‫اﻟ‬ ‫ﻟﻠﺼﻒ‬
‫ﺗــﺄﻟــﻴﻒ‬
‫اﻟﺤﺪﻳﺜﻲ‬ ‫رﺟﺐ‬ ‫ﺷﻌﺒﺎن‬ ‫ﻃﺎرق‬ / ‫اﻟﺪﻛﺘﻮر‬
‫اﻟﻤﻌﻤﺎر‬ ‫ﺷﺮﻳﻒ‬ ‫ﻳﻮﺳﻒ‬
‫اﻟﺠﻮاﻫﺮي‬ ‫اﻟﻐﻔﻮر‬ ‫ﻋﺒﺪ‬ ‫ﻣﺤﻤﺪ‬
‫ﻡ‬ ٢٠١٦ / ‫ﻫـ‬ ١٤٣٧ ‫اﻟﺘﺎﺳﻌﺔ‬ ‫اﻟﻄﺒﻌـﺔ‬

:‫مقدمة‬
‫لطلبة‬ ‫الجديدة‬ ‫الرياضيات‬ ‫كتب‬ ‫سلسلة‬ ‫في‬ ‫األولى‬ ‫الحلقة‬ ‫الكتاب‬ ‫هذا‬ ّ‫د‬‫ع‬ُ‫ي‬
‫هذا‬ ‫في‬ ً‫ا‬‫تخصص‬ ‫األكثر‬ ‫اسة‬‫ر‬‫الد‬ ‫بداية‬ ‫بوصفه‬ ‫العلمي‬ ‫للفرع‬ ‫اإلعدادية‬ ‫اسة‬‫ر‬‫الد‬
:‫وهي‬ ‫فصول‬ ‫سبعة‬ ‫على‬ ‫اشتمل‬ ‫وقد‬ ‫المجال‬
.‫الرياضي‬ ‫المنطق‬ ‫يتضمن‬ : ‫االول‬ ‫الفصــــل‬
.‫والمتباينات‬ ‫المعادالت‬ : ‫الثاني‬ ‫الفصـــل‬
.‫والجذور‬ ‫االسس‬ ‫في‬ ‫األساسية‬ ‫المبادئ‬ ‫تضمن‬ : ‫الثالث‬ ‫الفصــل‬
.‫المثلثات‬ ‫حساب‬ ‫في‬ ‫أساسية‬ ‫معلومات‬ ‫تضمن‬ : ‫ابع‬‫ر‬‫ال‬ ‫الفصـــل‬
‫هندسة‬ ‫مجال‬ ‫في‬ ‫األساسية‬ ‫المفاهيم‬ ‫الفصل‬ ‫هذا‬ ‫تضمن‬ : ‫الخامس‬ ‫الفصل‬
.‫المتجهات‬
‫الهندسة‬ ‫مجال‬ ‫في‬ ‫األساسية‬ ‫والمفاهيم‬ ‫المعلومات‬ ‫تضمن‬ : ‫السادس‬ ‫الفصل‬
.‫االحداثية‬
‫المرحلة‬ ‫في‬ ‫الطالب‬ ‫درسه‬ ‫لما‬ ً‫ال‬‫مكم‬ ‫الفصل‬ ‫هذا‬ ‫جاء‬ : ‫السابع‬ ‫الفصــل‬
.‫االحصاء‬ ‫مادة‬ ‫في‬ ‫المتوسطة‬
‫العزيز‬ ‫لبلدنا‬ ‫الخير‬ ‫فيه‬ ‫ما‬ ‫الى‬ ‫يوفقنا‬ ‫ان‬ ‫القدير‬ ‫العلي‬ ‫الله‬ ‫من‬ ‫نرجو‬ ‫الختام‬ ‫في‬
.‫الموفق‬ ‫والله‬ ‫التطوير‬ ‫بهدف‬ ‫بمالحظاتهم‬ ‫موافاتنا‬ ‫المدرسين‬ ‫زمالئنا‬ ‫من‬ ‫ونأمل‬
‫المؤلفون‬

1 ‫ﺍﻟﺮﻳﺎﺿﻲ‬ ‫ﺍﻟﻤﻨﻄﻖ‬ : ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﺼﻞ‬
‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬ [1-1]
‫....ﻓﺎن‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬ [1-2]
‫اذا‬ ‫ﻓﻘﻂ‬ ‫و‬ ‫اذا‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬ [1-3]
‫اﻻﻗﺘﻀﺎء‬ [1-4]
‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ [1-5]
‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﺗﻜﺎﻓﺆ‬ [1-6]
‫اﻟﻤﺴﻮرة‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ [1-7]
‫اﻟﺮﻳﺎﺿﻴﺔ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫او‬ ‫اﻟﺮﻣﺰ‬ ‫اﻟﻤﺼﻄﻠﺢ‬
‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬
∧ ‫و‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬
∨ ‫او‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬
← ‫ﻓﺄن‬ ... ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬
↔ ‫اذا‬ ‫وﻓﻘﻂ‬ ‫اذا‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬
⇔ ، ⇐ ‫اﻻﻗﺘﻀﺎء‬
∃ ،∀ ‫واﻟﺠﺰﺋﻲ‬ ‫اﻟﻜﻠﻲ‬ : ‫اﻟﺘﺴﻮﻳﺮ‬
‫اﻟﺴﻠﻮﻛﻴﺔ‬ ‫اﻻﻫﺪاف‬
:‫ان‬ ‫ﻋﻠﻰ‬ ً‫ا‬‫ر‬‫ﻗﺎد‬ ‫اﻟﻤﻮﺿﻮع‬ ‫ﻟﻬﺬا‬ ‫اﺳﺘﻪ‬‫ر‬‫د‬ ‫ﻧﻬﺎﻳﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻄﺎﻟﺐ‬ ‫ﻳﺼﺒﺢ‬ ‫ان‬ ‫ﻳﻨﺒﻐﻲ‬
‫اﻟﻤﺮﻛﺒﺔ‬ ‫ات‬‫ر‬‫واﻟﻌﺒﺎ‬ ‫وﻧﻔﻴﻬﺎ‬ ‫ات‬‫ر‬‫ﻟﻠﻌﺒﺎ‬ ‫اﻟﺼﻮاب‬ ‫ﻗﻴﻢ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ -
‫اﻟﺮﺑﻂ‬ ‫ادوات‬ ‫ﻋﻠﻰ‬ ‫اﻟﺘﻌﺮف‬ ‫ﺧﻼل‬ ‫ﻣﻦ‬ ‫اﻟﻤﺮﻛﺒﺔ‬ ‫واﻟﺠﻤﻞ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ -
‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﺗﻜﺎﻓﺆ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ -
‫وﻧﻔﻴﻬﺎ‬ ‫اﻟﻤﺴﻮرة‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ -

5
Mathematical logic ‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ : ‫اﻷول‬ ‫اﻟﻔﺼﻞ‬
‫ﻟﻜﻲ‬ ‫اﻵﺧﺮ‬ ‫ﺑﻌﻀﻬﺎ‬ ‫ﻋﻠﻰ‬ ‫ﺑﻌﻀﻬﺎ‬ ‫اﻟﻤﺮﺗﺒﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻣﻦ‬ ‫ﺳﻠﺴﻠﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫ﺗﺤﺘﺎج‬ � ‫ﺗﻤﻬﻴﺪ‬
. ‫ﻣﻨﻄﻘﻲ‬ ‫ﻧﻈﺎم‬ ‫أﻧﻬﺎ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫إﻟﻰ‬ ‫اﻟﻨﻈﺮ‬ ‫ﻳﻤﻜﻦ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻫﺬه‬ ‫وﻣﻦ‬ . ‫ﺻﺤﻴﺤﺔ‬ ‫ﻧﺘﺎﺋﺞ‬ ‫اﻟﻰ‬ ‫ﻧﺼﻞ‬
‫ﻳﺴﻤﻰ‬ ‫ﻣﺎ‬ ‫ن‬ّ‫ﻮ‬‫ﻳﻜ‬ ‫اﻻﺳﺘﺨﺪام‬ ‫ﺳﻬﻠﺔ‬ ‫ﺛﺎﺑﺘﺔ‬ ‫ﻗﻮاﻋﺪ‬ ‫وﺿﻊ‬ ‫ﻣﻊ‬ ‫رﻣﺰﻳﺔ‬ ‫ﺻﻮر‬ ‫ﻓﻲ‬ ‫اﻟﺮﻳﺎﺿﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫وﻛﺘﺎﺑﺔ‬
‫ﺑﻴﻦ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﻣﺘﻔﻖ‬ ‫ﻋﻠﻤﻴﺔ‬ ‫ﻟﻐﺔ‬ ‫ﻟﻜﻨﻪ‬ ‫ﻧﻈﺮﻳﺔ‬ ‫ﻟﻴﺲ‬ ‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ ‫ﻓﺎن‬ ‫وﻋﻠﻴﻪ‬ (‫اﻟﺮﻳﺎﺿﻲ‬ ‫)اﻟﻤﻨﻄﻖ‬
‫ﺣﺴﺐ‬ ‫ﻛﻞ‬ ‫ﻓﻬﻤﻬﺎ‬ ‫ﻓﻲ‬ ‫اء‬‫ﺮ‬‫اﻟﻘ‬ ‫ﻳﺨﺘﻠﻒ‬ ‫أن‬ ‫اﻟﺠﺎﺋﺰ‬ ‫ﻣﻦ‬ (‫)اﻟﺪارﺟﺔ‬ ‫اﻻﻋﺘﻴﺎدﻳﺔ‬ ‫ﻓﺎﻟﻠﻐﺔ‬ ‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫ﻋﻠﻤﺎء‬
، ‫اﻟﺨﻼف‬ ‫ﻟﻬﺬا‬ ‫اﻟﺮﻳﺎﺿﻴﺔ‬ (‫ات‬‫ر‬‫)اﻟﻌﺒﺎ‬ ‫ﻤﻞ‬ُ‫اﻟﺠ‬ ‫ﻣﻔﻬﻮم‬ ‫ﻧﺘﺮك‬ ‫أن‬ ‫ﻧﺴﺘﻄﻴﻊ‬ ‫ﻓﻼ‬ ‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫ﻓﻲ‬ ‫أﻣﺎ‬ ،‫ﻗﺪرﺗﻪ‬
.‫ﻧﺴﺘﻌﻤﻠﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺮﻳﺎﺿﻴﺔ‬ ‫ﻤﻞ‬ُ‫اﻟﺠ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻘﺼﻮد‬ ‫ﻟﺘﻔﺴﻴﺮ‬ ‫إﺗﻔﺎﻗﺎت‬ ‫اﻟﻌﻠﻤﺎء‬ ‫وﺿﻊ‬ ‫وﻟﺬا‬
Logical statement � ‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬ �1�1�
‫ﻧﻘﺴﻢ‬ ‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ ‫ﻣﻮﺿﻮع‬ ‫ﻓﻲ‬ ‫اﻟﻤﺘﻮﺳﻂ‬ ‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﺼﻒ‬ ‫ﻓﻲ‬ ‫درﺳﻨﺎ‬ ‫ﻟﻘﺪ‬ ‫اﻟﺪرس‬ ‫ﺗﻮاﺻﻞ‬
: ‫ﻧﻮﻋﻴﻦ‬ ‫اﻟﻰ‬ ‫ﻤﻞ‬ُ‫اﻟﺠ‬
. ً‫ﺎ‬‫ﻣﻌﻴﻨ‬ ً‫ا‬‫ﺮ‬‫ﺧﺒ‬ ‫إﻟﻴﻨﺎ‬ ‫ﺗﺤﻤﻞ‬ ‫ﻻ‬ ‫ﺟﻤﻠﺔ‬ ( ‫أ‬
. ( ‫ﺧﺒﺮﻳﺔ‬ ‫ﺟﻤﻠﺔ‬ ) ً‫ﺎ‬‫ﻣﻌﻴﻨ‬ ً‫ا‬‫ﺮ‬‫ﺧﺒ‬ ‫إﻟﻴﻨﺎ‬ ‫ﺗﺤﻤﻞ‬ ‫ﺟﻤﻠﺔ‬ (‫ب‬
‫وﻟﻘﺪ‬ . ‫ﺧﺎﻃﺌﺔ‬ ‫أو‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﺨﺒﺮﻳﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫ﻣﺎ‬ ‫ﻣﻌﺮﻓﺔ‬ ‫ﻫﻮ‬ ‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ ‫ﻣﻬﺎم‬ ‫ﻣﻦ‬ ‫وان‬
‫ﺗﻜﻮن‬ ‫أن‬ ‫ﻳﻤﻜﻦ‬ ‫وﻻ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫أو‬ ‫ﺻﺎﺋﺒﺔ‬ ‫إﻣﺎ‬ ‫وﻫﻲ‬ ‫ﻣﻨﻄﻘﻴﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﺗﺴﻤﻰ‬ ‫اﻟﺨﺒﺮﻳﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫أن‬ ‫ﻋﺮﻓﺖ‬
. ‫واﺣﺪ‬ ‫وﻗﺖ‬ ‫ﻓﻲ‬ ‫وﺧﺎﻃﺌﺔ‬ ‫ﺻﺎﺋﺒﺔ‬
( 1 - 1 ) ‫الجدول‬
‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻣﻨﻄﻘﻴﺔ‬ ‫ﻟﻌﺒــﺎرة‬ ‫رﻣﺰﻧـــﺎ‬ ‫إذا‬ ‫اﻧﻪ‬ ‫ﻋﻠﻤﺖ‬ ‫وﻟﻘﺪ‬
‫اذا‬ (T) (True) ‫ﺻــﺎﺋﺒﺔ‬ ‫ﺗﻜــﻮن‬ P ‫ﻧﻔﻲ‬ ‫ﻓﺎن‬ P
P ‫ﻧﻔـــﻲ‬ ‫وﻳﻜــﻮن‬ (F) (False) ‫ﺧﺎﻃﺌﺔ‬ P ‫ﻛﺎﻧﺖ‬
‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫ذﻟﻚ‬ ‫ﻋﻦ‬ ‫وﻋﺒﺮﻧﺎ‬ ‫ﺻﺎﺋﺒﺔ‬ P ‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫ﺧﺎﻃﺌﺔ‬
. ( 1 - 1 ) ‫اﻟﺠﺪول‬
∼ P P
F T
T F

6
:(∨) ‫او‬ ، (∧) ‫و‬ ‫اﻟﺮﺑﻂ‬ ‫ﻷداﺗﻲ‬ ‫اﻟﺼﻮاب‬ ‫ﺟﺪوﻟﻲ‬ ‫ﻧﺬﻛﺮ‬ ‫ان‬ ‫اﻟﻤﻔﻴﺪ‬ ‫وﻣﻦ‬
(1-3) ‫الجدول‬ (1-2) ‫الجدول‬
If �� then ( ‫ﻓﺄن‬ �� ‫ﻛﺎن‬ ‫إذا‬ � � ‫اﻟﺮﺑﻂ‬ ‫أداة‬ �1�2�
‫ﻟﺘﻜﻮﻳﻦ‬ ‫ﺗﺴﺘﺨﺪم‬ ‫اﻟﺘﻲ‬ ‫اﻟﺮواﺑﻂ‬ ‫اﺣﺪى‬ ‫ﻓﻬﻲ‬ ( ‫ﻓﺎن‬ ... ‫ﻛﺎن‬ ‫)إذا‬ ‫اﻟﺮﺑﻂ‬ ‫اداة‬ ‫ﻋﻠﻰ‬ ‫اﻻن‬ ‫ﺳﻨﺘﻌﺮف‬
.( Compound Statement ) ‫اﻟﻤﺮﻛﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬
:‫اﻟﻤﺮﻛﺒﺔ‬ ‫ﻓﺎﻟﻌﺒﺎرة‬
‫ﻣﻦ‬ ‫ﺗﻜﻮﻧﺖ‬ ‫ﻣﺘﺴﺎوﻳﺎن‬ ‫ﻗﺎﻋﺪﺗﻪ‬ ‫اوﻳﺘﻲ‬‫ز‬ ‫ﻗﻴﺎﺳﻲ‬ ‫ﻓﺄن‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﺟـ‬ ‫ب‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻛﺎن‬ ‫اذا‬
:‫اﻟﻌﺒﺎرة‬ ‫رﺑﻂ‬
((‫ﻣﺘﺴﺎوﻳﺎن‬ ‫ﺟـ‬ ‫ب‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻗﺎﻋﺪة‬ ‫اوﻳﺘﻲ‬‫ز‬ ‫))ﻗﻴﺎﺳﺎ‬ ‫ﺑﺎﻟﻌﺒﺎرة‬ ((‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﺟـ‬ ‫ب‬ ‫))اﻟﻤﺜﻠﺚ‬
. ((‫ﻓﺎن‬ .... ‫ﻛﺎن‬ ‫))اذا‬ . ‫اﻟﺮﺑﻂ‬ ‫ﺑﺄداة‬
(‫)ﻓﺈن‬ ‫ﺑﻌﺪ‬ ‫ﺗﺄﺗﻲ‬ ‫اﻟﺘﻲ‬ ‫واﻟﻌﺒﺎرة‬ ‫ﺑﺎﻟﻤﻘﺪﻣﺔ‬ (‫ﻛﺎن‬ ‫)اذا‬ ‫ﺗﻠﻲ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﺗﺴﻤﻴﺔ‬ ‫ﻋﻠﻰ‬ ‫اﺻﻄﻠﺢ‬ ‫وﻗﺪ‬
: ‫اﻻداة‬ ‫ﺗﺴﻤﻰ‬ ‫ﻛﻤﺎ‬ . ‫ﺑﺎﻟﺘﺎﻟﻴﺔ‬
.‫اﻟﺸﺮط‬ ‫أداة‬ ( ‫ﻓﺎن‬ .... ‫ﻛﺎن‬ ‫)إذا‬
‫ﻛﻤﺎ‬ ،ً‫ﺎ‬‫ﺳﺎﺑﻘ‬ ‫اﻟﻤﺬﻛﻮرة‬ ‫اﻟﻤﺮﻛﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻣﻘﺪﻣﺔ‬ ‫ﺗﻜﻮن‬ (‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﺟـ‬ ‫ب‬ ‫)اﻟﻤﺜﻠﺚ‬ ‫ﻓﺎﻟﻌﺒﺎرة‬
. ‫ﺗﺎﻟﻴﻬﺎ‬ ‫ﻫﻲ‬ (‫ﻣﺘﺴﺎوﻳﺎن‬ ‫ﺟـ‬ ‫ب‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻗﺎﻋﺪة‬ ‫اوﻳﺘﻲ‬‫ز‬ ‫)ﻗﻴﺎﺳﺎ‬ ‫اﻟﻌﺒﺎرة‬ ‫أن‬
P Q P ∧ Q
T T T
T F F
F T F
F F F
P Q P ∨ Q
T T T
T F T
F T T
F F F

7
: ‫اﻵتي‬ ‫المثال‬ ‫لنأﺧذ‬ ‫واالن‬
(‫ﻫﺪﻳﺔ‬ ‫ﻓﺴﺄﻋﻄﻴﻚ‬ ‫اﻻﻣﺘﺤﺎن‬ ‫ﻓﻲ‬ ‫ﻧﺠﺤﺖ‬ ‫)اذا‬ : ‫ﻟﻮﻟﺪﻫﺎ‬ ‫اﻷم‬ ‫ﻗﺎﻟﺖ‬
:‫اﻵﺗﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻟﻨﺪرس‬
‫ﻫﺪﻳﺔ‬ ‫أﻣﻪ‬ ‫ﻟﻪ‬ ‫وﻗﺪﻣﺖ‬ ‫اﻻﻣﺘﺤﺎن‬ ‫ﻓﻲ‬ ‫اﻟﻮﻟﺪ‬ ‫ﻧﺠﺢ‬ (1
‫ﻫﺪﻳﺔ‬ ‫أﻣﻪ‬ ‫ﻟﻪ‬ ‫ﺗﻘﺪم‬ ‫وﻟﻢ‬ ‫اﻻﻣﺘﺤﺎن‬ ‫ﻓﻲ‬ ‫اﻟﻮﻟﺪ‬ ‫ﻧﺠﺢ‬ (2
‫ﻫﺪﻳﺔ‬ ‫أﻣﻪ‬ ‫ﻟﻪ‬ ‫وﻗﺪﻣﺖ‬ ‫اﻻﻣﺘﺤﺎن‬ ‫ﻓﻲ‬ ‫اﻟﻮﻟﺪ‬ ‫ﻳﻨﺠﺢ‬ ‫ﻟﻢ‬ (3
‫ﻫﺪﻳﺔ‬ ‫أﻣﻪ‬ ‫ﻟﻪ‬ ‫ﺗﻘﺪم‬ ‫وﻟﻢ‬ ‫اﻻﻣﺘﺤﺎن‬ ‫ﻓﻲ‬ ‫اﻟﻮﻟﺪ‬ ‫ﻳﻨﺠﺢ‬ ‫ﻟﻢ‬ (4
‫ﺣﺼﻠﺖ‬ ‫إذا‬ ‫أﻣﺎ‬ ‫اﺑﻌﺔ‬‫ﺮ‬‫واﻟ‬ ‫واﻟﺜﺎﻟﺜﺔ‬ ‫اﻷوﻟﻰ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫اﻷم‬ ‫ذﻛﺮﺗﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﺻﻮاب‬ ‫ﻧﻘﺒﻞ‬ ‫ﺳﻮف‬
. ‫ﺧﺎﻃﺌﺔ‬ ‫ﺗﻜﻮن‬ ‫ذﻛﺮﺗﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻓﺈن‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺤﺎﻟﺔ‬
( ‫....ﻓﺈن‬ ‫ﻛﺎن‬ ‫)إذا‬ ‫اﻟﺮﺑﻂ‬ ‫ﻷداة‬ ‫ﻣﺤﺪد‬ ‫اﺳﺘﻌﻤﺎل‬ ‫ﻋﻠﻰ‬ ‫وﺳﻨﺘﻔﻖ‬
. P → Q ‫ﺑﺎﻟﺮﻣﺰ‬ :‫اﻟﻤﺮﻛﺒﺔ‬ ‫ﻟﻠﻌﺒﺎرة‬ ‫ﻳﺮﻣﺰ‬ ‫ﻓﺈﻧﻪ‬ ‫ﻋﺒﺎرﺗﻴﻦ‬ Q ، P ‫ﻛﺎﻧﺖ‬ ‫ﻓﺈذا‬
(( Q ‫ﻓﺎن‬ P ‫ﻛﺎن‬ ‫اذا‬ )) ‫وﺗﻘﺮأ‬
: ‫ﻛﺎﻵﺗﻲ‬ P → Q ‫ﻟﻠﻌﺒﺎرة‬ ‫اﻟﺼﻮاب‬ ‫ﺟﺪول‬ ‫ﻳﻜﻮن‬ ‫أن‬ ‫ﻋﻠﻰ‬ ‫اﺗﻔﻖ‬ ‫وﻗﺪ‬
(1-4) ‫الجدول‬
. ‫فقﻂ‬ « ‫ﺧاﻃﺌة‬ » ‫والتالية‬ « ‫»صائبة‬ ‫المقدمة‬ ‫كانﺖ‬ ‫إذا‬ ‫ﺧاﻃﺌة‬ ‫تكون‬ P →Q ‫أن‬ ‫أي‬
P → QQP
TTT
FFT
TTF
TFF

8
: ‫اﻵﺗﻴﺔ‬ ‫ات‬‫ر‬‫ﻟﻠﻌﺒﺎ‬ ‫اﻟﺼﻮاب‬ ‫ﻗﻴﻢ‬ ‫ذﻛﺮ‬ُ‫ا‬ � � 1 ‫ﻣﺜﺎل‬
‫ﻓﺎن‬ 2 < 3 ‫ﻛﺎن‬ ‫اذا‬ (1
7 =6+2 ‫ﻓﺈن‬ 12 = 7+5 ‫ﻛﺎن‬ ‫إذا‬ (2
8=6+2 ‫ﻓﺈن‬ 11= 7+5 ‫ﻛﺎن‬ ‫إذا‬ (3
‫ﻧﺴﺒﻲ‬ ‫ﻋﺪد‬ 3 ‫ﻓﺈن‬ 1= ‫ﺻﻔﺮ‬ ‫ﻛﺎن‬ ‫إذا‬ (4
� ‫اﻟﺤــــﻞ‬
.‫ﺻﺎﺋﺒﺔ‬ ‫واﻟﺘﺎﻟﻴﺔ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﻤﻘﺪﻣﺔ‬ ‫ﻷن‬ ‫ﺻﺎﺋﺒﺔ‬ (1
.‫ﺧﺎﻃﺌﺔ‬ ‫واﻟﺘﺎﻟﻴﺔ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﻤﻘﺪﻣﺔ‬ ‫ﻷن‬ ‫ﺧﺎﻃﺌﺔ‬ (2
.‫ﺻﺎﺋﺒﺔ‬ ‫واﻟﺘﺎﻟﻴﺔ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫اﻟﻤﻘﺪﻣﺔ‬ ‫ﻷن‬ ‫ﺻﺎﺋﺒﺔ‬ (3
.‫ﺧﺎﻃﺌﺔ‬ ‫واﻟﺘﺎﻟﻴﺔ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫اﻟﻤﻘﺪﻣﺔ‬ ‫ﻷن‬ ‫ﺻﺎﺋﺒﺔ‬ (4
If and only if (‫إذا‬ ‫وﻓﻘﻂ‬ ‫�إذا‬ � ‫اﻟﺮﺑﻂ‬ ‫أداة‬ [1�3�
:‫اﻟﻤﺮﻛﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻧﺴﺘﻌﻤﻞ‬ ‫ﻣﺎ‬ ً‫ا‬‫ﺮ‬‫ﻛﺜﻴ‬
(Q → P) ∧(P → Q)
‫ﻗﻴﺎﺳﺎت‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫ﻛﺬﻟﻚ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬ ‫زواﻳﺎه‬ ‫ﻗﻴﺎﺳﺎت‬ ‫ﻓﺈن‬ ‫اﻻﺿﻼع‬ ‫ﻣﺘﺴﺎوي‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻛﺎن‬ ‫إذا‬ : ً‫ﻼ‬‫ﻓﻤﺜ‬
‫ﺷﺮﻃﻴﺔ‬ ‫)ﻋﺒﺎرة‬ ‫اﻟﻤﺮﻛﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻫﺬه‬ ‫أﻣﺜﺎل‬ ‫ﺗﺴﻤﻰ‬ . ‫اﻻﺿﻼع‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻳﻜﻮن‬ ‫ﻓﺎﻧﻪ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬ ‫ﻣﺜﻠﺚ‬ ‫زواﻳﺎ‬
‫اﻟﺜﻨﺎﺋﻴﺔ‬ ‫اﻟﺸﺮﻃﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻓﺈن‬ ‫ﻋﺒﺎرﺗﻴﻦ‬ P،Q ‫ﻓﺮﺿﻨﺎ‬ ‫ﻓﺈذا‬ ( ‫ﺛﻨﺎﺋﻴﺔ‬
P↔Q ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻬﺎ‬ ‫ﻳﺮﻣﺰ‬ (Q → P) ∧ (P → Q)
(Q ‫إذا‬ ‫وﻓﻘﻂ‬ ‫إذا‬ P) : ‫وﺗﻘﺮأ‬
P↔Q : ‫اﻟﻌﺒﺎرة‬ ‫ﺻﻮاب‬ ‫ﺟﺪول‬ ‫ﻫﻮ‬ ( 1 - 5 ) ‫واﻟﺠﺪول‬
-2∉ R

9
(1-5) ‫جدول‬
‫ﺻﺎﺋﺒﺘﻴﻦ‬ ‫ﻟﻬﺎ‬ ‫اﻟﻤﺮﻛﺒﺘﻴﻦ‬ ‫اﻟﻌﺒﺎرﺗﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ : ‫ﻫﻤﺎ‬ ‫ﺣﺎﻟﺘﻴﻦ‬ ‫ﻓﻲ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫ﺗﻜﻮن‬ P ↔ Q ‫أن‬ ‫أي‬
.ً‫ﺎ‬‫ﻣﻌ‬ ‫ﺧﺎﻃﺌﺘﻴﻦ‬ ‫أو‬ ً‫ﺎ‬‫ﻣﻌ‬
� � 2 ‫ﻣﺜﺎل‬
X=-1,X=4 ↔ X2
-3X-4=0 ( ‫أ‬
X5
=-32 ↔ X=-2 (‫ب‬
Implication ‫اﻻﻗﺘﻀﺎء‬ [1�4�
: ‫اﻵﺗﻴﺘﻴﻦ‬ ‫اﻟﺤﺎﻟﺘﻴﻦ‬ ‫ﺧﻼل‬ ‫ﻣﻦ‬ ‫اﻻﻗﺘﻀﺎء‬ ‫ﻣﻌﻨﻰ‬ ‫ﺳﻨﻮﺿﺢ‬
⇒ ‫ﻟﻪ‬ ‫ﺮﻣﺰ‬ُ‫ﻳ‬ ‫واﻟﺬي‬ ‫واﺣﺪ‬ ‫اﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫اﻻﻗﺘﻀﺎء‬ : ‫اﻷوﻟﻰ‬ ‫اﻟﺤﺎﻟﺔ‬
P ‫ﺑﺎﻟﺮﻣﺰ‬ «X=3» : ‫ﻟﻨﺮﻣﺰ‬
Q ‫ﺑﺎﻟﺮﻣﺰ‬ «X2
= 9» : ‫وﻟﻨﺮﻣﺰ‬
X2
=9 ‫ﺗﻜﻮن‬ ‫أن‬ ‫ﻳﻘﺘﻀﻲ‬ ‫ﻫﺬا‬ ‫ﻓﺈن‬ ‫ﺻﺎﺋﺒﺔ‬ X = 3 ‫ﻛﺎﻧﺖ‬ ‫ﻓﺈذا‬
P ⇒ Q : ‫اي‬
X = ± 3 ‫ﻓﺎن‬ X2
=9 ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫أﻣﺎ‬
Q ⇏ P : ‫اي‬
P Q P → Q Q → P (p → Q)∧(Q →p)
T T T T T
T F F T F
F T T F F
F F T T T
P↔Q

10
⇔ ‫ﻟﻪ‬ ‫ﺮﻣﺰ‬ُ‫ﻳ‬ ‫واﻟﺬي‬ ‫ﻣﺘﻌﺎﻛﺴﻴﻦ‬ ‫اﺗﺠﺎﻫﻴﻦ‬ ‫ﻓﻲ‬ ‫اﻻﻗﺘﻀﺎء‬ : ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺤﺎﻟﺔ‬
P ‫ﺑﺎﻟﺮﻣﺰ‬ «X =3» ‫ﻟﻨﺮﻣﺰ‬
Q ‫ﺑﺎﻟﺮﻣﺰ‬ «X3
=27» ‫وﻟﻨﺮﻣﺰ‬
X3
=27 ‫ﺗﻜﻮن‬ ‫أن‬ ‫ﻳﻘﺘﻀﻲ‬ ‫ﻫﺬا‬ ‫ﻓﺈن‬ ‫ﺻﺎﺋﺒﺔ‬ X = 3 ‫ﻛﺎﻧﺖ‬ ‫ﻓﺈذا‬
P ⇒ Q ‫أي‬
X = 3 ‫ﺗﻜﻮن‬ ‫أن‬ ‫ﻳﻘﺘﻀﻲ‬ ‫ﻫﺬا‬ ‫ﻓﺈن‬ ‫ﺻﺎﺋﺒﺔ‬ X3
= 27 ‫ﻛﺎﻧﺖ‬ ‫وإذا‬
Q⇒ P ‫أي‬
P ⇔ Q ‫ان‬ ‫ﻳﻌﻨﻲ‬ (Q⇒ P) ∧ (P⇒ Q) ‫أن‬
� � 3 ‫ﻣﺜﺎل‬
. ‫ﺻﺤﻴﺤﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻟﺘﺼﺒﺢ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫اﻟﺘﻌﺒﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫ﻟﻮﺿﻌﻪ‬ ⇔ ، ⇐ ‫اﻟﺮﻣﺰﻳﻦ‬ ‫أﺣﺪ‬ ‫إﺧﺘﺮ‬
X3
=8 ( ‫أ‬
X>2 , X>5 (‫ب‬
X2
≥0 , X≤0 (‫ﺟـ‬
‫اﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫د‬ ‫ﺟـ‬ ‫ب‬ : Q ، ‫ﻣﺘﻨﺎﺻﻔﺎن‬ ‫اه‬‫ﺮ‬‫ﻗﻄ‬ ‫رﺑﺎﻋﻲ‬ ‫ﺷﻜﻞ‬ ‫د‬ ‫ﺟـ‬ ‫ب‬ : P (‫د‬
X3
=8 ⇔ X=2 ( ‫أ‬
X>5 ⇒ X>2 (‫ب‬
X≤0 ⇒ X2
≥ 0(‫ﺟـ‬
Q ⇔ P (‫د‬
X = 2 ,

11
Equivalent Statements ‫اﻟﻤﺘﻜﺎﻓﺌﺘﺎن‬ ‫اﻟﻌﺒﺎرﺗﺎن‬
(1-1) ‫ﺗﻌﺮﻳﻒ‬
P≡Q‫ﺑﺎﻟﺮﻣﺰ‬‫ﻟﻬﺎ‬‫ﻟﻠﻌﺒﺎرةوﻳﺮﻣﺰ‬‫اﻟﺼﻮاب‬‫ﺟﺪول‬‫ﻧﻔﺲ‬‫ﻟﻬﺎ‬‫ﻛﺎن‬‫اذا‬Q‫ﻟﻠﻌﺒﺎرة‬‫ﻣﻜﺎﻓﺌﺔ‬P‫اﻟﻌﺒﺎرة‬‫أن‬‫ﻳﻘﺎل‬
� � 4 ‫ﻣﺜﺎل‬
P→Q≡~ P∨Q ‫أن‬ ‫أﺛﺒﺖ‬
: ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻧﻌﻤﻞ‬
P Q ~ P P →Q ~P ∨Q
T T F T T
T F F F F
F T T T T
F F T T T
≡

12
( 1 - 1 ) ‫تمرينات‬
� 1‫س‬
: ‫اﻟﺴﺒﺐ‬ ‫ﻣﻊ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫ﻣﻨﻬﺎ‬ ً‫ﺎ‬‫وأﻳ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﺘﺎﻟﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻣﻦ‬ ًّ‫ﺎ‬‫أﻳ‬ ‫ﺑﻴﻦ‬
. 25 ‫ﻘﺴﻢ‬ُ‫ﻳ‬ 7 ‫واﻟﻌﺪد‬ 25 ‫اﻟﻌﺪد‬ ‫ﻘﺴﻢ‬ُ‫ﻳ‬ 5 ‫اﻟﻌﺪد‬ ( ‫أ‬
. 25 ‫ﻘﺴﻢ‬ُ‫ﻳ‬ 7 ‫اﻟﻌﺪد‬ ‫أو‬ 25 ‫اﻟﻌﺪد‬ ‫ﻘﺴﻢ‬ُ‫ﻳ‬ 5 ‫اﻟﻌﺪد‬ (‫ب‬
. ً‫ﺎ‬‫أوﻟﻴ‬ 4 ‫اﻟﻌﺪد‬ ‫أو‬ ً‫ﺎ‬‫أوﻟﻴ‬ ‫ﻟﻴﺲ‬ 7 ‫اﻟﻌﺪد‬ (‫ﺟـ‬
. ‫ﻣﺘﻨﺎﺻﻔﺎن‬ ‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ا‬‫ﺮ‬‫ﻗﻄ‬ ‫و‬ ‫ﻣﺘﻌﺎﻣﺪان‬ ‫اﻟﻤﺮﺑﻊ‬ ‫ا‬‫ﺮ‬‫ﻗﻄ‬ (‫د‬
. ‫ﻣﺘﻌﺎﻣﺪان‬ ‫اﻟﻤﺴﺘﻄﻴﻞ‬ ‫ا‬‫ﺮ‬‫ﻗﻄ‬ ‫أو‬ ‫ﻣﺘﻌﺎﻣﺪان‬ ‫اﻟﻤﺮﺑﻊ‬ ‫ا‬‫ﺮ‬‫ﻗﻄ‬ (‫ﻫـ‬
� 2‫س‬
‫اﻟﻨﺎﺗﺠﺔ‬ ‫اﻟﻤﺮﻛﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﺗﺼﺒﺢ‬ ‫ﻟﻜﻲ‬ ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻓﻲ‬ ‫اﻟﻌﺒﺎرﺗﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫ﻟﻠﺮﺑﻂ‬ ⇐ ‫أو‬ ⇔ ‫اﺳﺘﺨﺪم‬
: ‫ﺻﺎﺋﺒﺔ‬
P ‫اﻟﻌﺒﺎرة‬‫اﻟﺮﻣﺰ‬Q ‫اﻟﻌﺒﺎرة‬
‫ﻣﺴﺘﻄﻴﻞ‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬‫ﻳﺘﻨﺎﺻﻔﺎن‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬ ‫ا‬‫ﺮ‬‫ﻗﻄ‬
‫ﻣﻌﻴﻦ‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬‫ﻣﺘﻄﺎﺑﻘﺔ‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬ ‫أﺿﻼع‬
‫ﻣﺴﺘﻄﻴﻞ‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬‫ﻗﻮاﺋﻢ‬ ‫زواﻳﺎه‬ ‫ﻗﻴﺎس‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬
a.b=0 , a,b∈Ra=0 ∨ b=0
X = -3X2
= 9
‫ﻣﺮﺑﻊ‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬‫ﻗﻮاﺋﻢ‬ ‫زواﻳﺎه‬ ‫ﻗﻴﺎس‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬
X2
= 25X = 5
X3
= -125X = -5
‫اﻻﺿﻼع‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬ ‫ﺟـ‬ ‫ب‬‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬ ‫ﺟـ‬ ‫ب‬
X=1 ∨ X=2(X-1)(X-2)=0

13
� 3‫س‬
: ‫ان‬ ‫ﺑﺮﻫﻦ‬
P→Q ≡~Q →~P ( 1
( 2
� 4‫س‬
‫؟‬ ‫ﺻﺎﺋﺒﺔ‬ ‫وأﻳﻬﺎ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫اﻵﺗﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻓﺄي‬ ‫ﺧﺎﻃﺌﺔ‬ S ، ‫ﺻﺎﺋﺒﺔ‬ Q ، ‫ﺻﺎﺋﺒﺔ‬ P ‫ﻛﺎﻧﺖ‬ ‫اذا‬
( 1
( 2
( 3
( 4
� 5 ‫س‬
- : ‫ﻳﻠﻲ‬ ‫ﻓﻴﻤﺎ‬ ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻻﺟﺎﺑﺔ‬ ‫رﻣﺰ‬ ‫ﺣﻮل‬ ‫داﺋﺮة‬ ‫ﺿﻊ‬
: ‫اﻻﺗﻴﺔ‬ ‫اﻻﺳﺌﻠﺔ‬ ‫ﻓﻲ‬ ‫اﻋﺘﻤﺪت‬ ‫ﻋﺒﺎرﺗﻴﻦ‬ S ، P
‫ﺗﻜﺎﻓﻰء‬ P→ ~P( 1
~P ∧P ( ‫د‬ ~P ( ‫ﺟـ‬ ~ P →P ( ‫ب‬ P →P ( ‫أ‬
‫ﻋﺒﺎرة‬ S ↔S ( 2
‫واﺣﺪة‬ ‫ﻣﺮة‬ ‫( ﺧﺎﻃﺌﺔ‬ ‫د‬ ً‫ﺎ‬‫داﺋﻤ‬ ‫ﺧﺎﻃﺌﺔ‬ ( ‫ﺟـ‬ ‫واﺣﺪة‬ ‫ﻣﺮة‬ ‫ﺻﺎﺋﺒﺔ‬ ( ‫ب‬ ً‫ﺎ‬‫داﺋﻤ‬ ‫ﺻﺎﺋﺒﺔ‬ ( ‫أ‬
- : ‫ﻫﻮ‬ ~S∨« 9 > 5 + 3 » ‫اﻟﻌﺒﺎرة‬ ‫ﻧﻔﻲ‬ ( 3
~S∨« 9 < 5 + 3 » ( ‫ب‬ ~S∨« 9 ≥ 5 + 3 » ( ‫أ‬
S∧« 9 ≤ 5 + 3 » ( ‫د‬ ~S∧« 9 ≤ 5 + 3 » (‫ﺟـ‬
~(P→Q) ≡P∧~Q
(P→Q) ∨S
(P↔S) ∧P
(S→Q) ∧P
(S↔S) ∨S

14
Open Sentences ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ [1�5�
. (ً‫ﺎ‬‫ﻣﻌ‬ ‫اﻻﺛﻨﺎن‬ ‫)وﻟﻴﺲ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫أو‬ ‫ﺻﺎﺋﺒﺔ‬ ‫إﻣﺎ‬ ‫ﺧﺒﺮﻳﺔ‬ ‫ﺟﻤﻠﺔ‬ ‫ﺑﺄﻧﻬﺎ‬ ‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻋﺮﻓﻨﺎ‬
: ‫اﻵﺗﻴﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﻻﺣﻈﻨﺎ‬ ‫اذا‬ ‫وﻟﻜﻦ‬
‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻬﺎ‬ ‫ﻧﺮﻣﺰ‬ ‫واﻟﺘﻲ‬ ‫اﻟﺼﻔﺮ‬ ‫ﻣﻦ‬ ‫أﻛﺒﺮ‬ ‫ﺻﺤﻴﺢ‬ ‫ﻋﺪد‬ X ( ‫أ‬
‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻬﺎ‬ ‫ﻧﺮﻣﺰ‬ ‫واﻟﺘﻲ‬ (‫ب‬
‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻬﺎ‬ ‫ﻧﺮﻣﺰ‬ ‫واﻟﺘﻲ‬ ‫ﺻﺤﻴﺤﺔ‬ ‫أﻋﺪاد‬ a ، b ‫ﺣﻴﺚ‬ (‫ﺟـ‬
.‫اق‬‫ﺮ‬‫اﻟﻌ‬ ‫ﻣﺪن‬ ‫أﺣﺪى‬ . . . . (‫د‬
‫ﻓﻲ‬ ‫ﻋﻮﺿﻨﺎ‬ ‫إذا‬ ‫وﻟﻜﻦ‬ . ‫ﻣﻨﻄﻘﻴﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺠﻤﻞ‬ ‫ﻫﺬه‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫أن‬ ‫اﻟﻘﻮل‬ ‫ﺑﺎﻻﻣﻜﺎن‬ ‫ﻟﻴﺲ‬ ‫وﺟﺪﻧﺎ‬
‫اﻋﻂ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫وﻫﺬه‬ (‫اﻟﺼﻔﺮ‬ ‫ﻣﻦ‬ ‫أﻛﺒﺮ‬ ‫ﺻﺤﻴﺢ‬ ‫)9ﻋﺪد‬ ‫ﺗﺼﺒﺢ‬ X ‫اﻟﺤﺮف‬ ‫ﺑﺪل‬ 9 ‫ﺑﺎﻟﻌﺪد‬ (‫)أ‬ ‫اﻟﺠﻤﻠﺔ‬
3 ‫ﺗﺴﺎوي‬ ‫ﻗﻴﻤﺔ‬ a ، b ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫أﻋﻄﻴﺖ‬ ‫وﻟﻮ‬ . ‫ﺧﺎﻃﺌﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻟﺘﺠﻌﻠﻬﺎ‬ (‫)ب‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻓﻲ‬ (Y) ‫ﻟـ‬ ‫ﻗﻴﻤﺔ‬
(‫)د‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﻨﺎﺳﺐ‬ ‫اغ‬‫ﺮ‬‫اﻟﻔ‬ ‫ﻓﻲ‬ ‫اﻻﺳﻢ‬ ‫ﺿﻊ‬ . ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫وﻫﻲ‬ (6=3+3) ‫اﻟﻌﺒﺎرة‬ ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬
. ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻟﺘﺠﻌﻠﻬﺎ‬
(1-2) ‫ﺗﻌﺮﻳﻒ‬
‫ﻟﺬﻟﻚ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻔﺮوﺿﺔ‬ ‫اﻻﺷﻴﺎء‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬ ً‫ﺎ‬‫ﻗﻴﻤ‬ ‫ﻳﺄﺧﺬ‬ ‫رﻣﺰ‬ ‫ﻫﻮ‬ ‫اﻟﻤﺘﻐﻴﺮ‬ (1
. ‫اﻟﻤﺘﻐﻴﺮ‬
‫ﻣﺘﻐﻴﺮ‬ ‫ﻛﻞ‬ ‫إﻋﻄﺎء‬ ‫ﻋﻨﺪ‬ ‫ﻋﺒﺎرة‬ ‫إﻟﻰ‬ ‫وﺗﺘﺤﻮل‬ ‫أﻛﺜﺮ‬ ‫أو‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺘﻮي‬ ‫ﺟﻤﻠﺔ‬ ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ (2
. ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻣﻦ‬ ‫ﻣﻌﻴﻨﺔ‬ ‫ﻗﻴﻤﺔ‬
� ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﺗﻜﺎﻓﺆ‬ [1�6�
:‫ﻟﺘﻜﻦ‬
‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫(ﻧﻼﺣﻆ‬Z) ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫وﻟﺘﻜﻦ‬
.{2} ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫ﻟﻠﺠﻤﻠﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫وإن‬ {2} ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫ﻟﻠﺠﻤﻠﺔ‬
. ‫ﻣﻨﻬﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺘﻲ‬ ‫ﻟﺘﺴﺎوي‬ ‫وذﻟﻚ‬ ‫ﻣﺘﻜﺎﻓﺌﺘﻴﻦ‬ ‫اﻟﻤﻔﺘﻮﺣﺘﺎن‬ ‫اﻟﺠﻤﻠﺘﺎن‬ ‫ﺗﺴﻤﻰ‬
P(X)
Y+1=3Q(Y)
a+b=6G(a,b)
P(X): 2X=4
Q(x): X-1=1
Q(X) P(X)
Q(X),P(X)

15
� � 5 ‫ﻣﺜﺎل‬
‫ﻛﺎﻧﺖ‬ ‫إذا‬
. Z ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻟﻜﻞ‬ ‫وﻣﺠﻤﻮﻋﺔاﻟﺘﻌﻮﻳﺾ‬
‫؟‬ ‫ﻣﺘﻜﺎﻓﺌﺘﺎن‬ ‫ﻫﻞ‬
‫ﻟﻠﺠﻤﻠﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫وأن‬ {2} ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫ﻟﻠﺠﻤﻠﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫ﻧﻼﺣﻆ‬
، ‫اﻟﻤﻔﺘﻮﺣﺘﻴﻦ‬ ‫اﻟﺠﻤﻠﺘﻴﻦ‬ ‫أن‬ ‫ﻧﻘﻮل‬ ‫ﻟﺬا‬ { } ≠ {2} ‫أن‬ ‫وﺑﻤﺎ‬ { } ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬
. ‫ﻣﺘﻜﺎﻓﺌﺘﻴﻦ‬ ‫ﻏﻴﺮ‬ ‫ﺟﻤﻠﺘﺎن‬
(1-3) ‫ﺗﻌﺮﻳﻒ‬
‫ﺟﻤﻠﺔ‬ ‫أي‬ ‫أو‬ « ً‫ﺎ‬‫ﺻﺤﻴﺤ‬ ‫»ﻟﻴﺲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻫﻲ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻧﻔﻲ‬ ‫إن‬
. ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻧﻔﻲ‬ ‫ﻋﻦ‬ ‫ﻟﻠﺘﻌﺒﻴﺮ‬ ‫اﻟﺮﻣﺰ‬ ‫ﻧﺴﺘﻌﻤﻞ‬ ‫وﺳﻮف‬ ‫ذﻟﻚ‬ ‫ﺗﻜﺎﻓﻲء‬ ‫ﻣﻔﺘﻮﺣﺔ‬
� � 6 ‫ﻣﺜﺎل‬
. Z ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ ‫ﻳﻠﻲ‬ ‫ﻓﻴﻤﺎ‬ ‫ﻣﻔﺘﻮﺣﺔ‬ ‫ﺟﻤﻠﺔ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫ﻟﻨﻔﺮض‬
Q(X):X2
=4
P(X):X=2
P(X),Q(X)
P(X)
Q(X)2,-22,-2P(X)
Q(X)
P(X)P(X)
~ P(X)P(X)
‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﻧﻔﻴﻬﺎ‬
X2
-4=0
‫زوﺟﻲ‬ ‫ﺻﺤﻴﺢ‬ ‫ﻋﺪد‬ X
X=4 ‫و‬ X+1≠6
X2
-4≠0
ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ﺎ‬‫ﺻﺤﻴﺤ‬ ً‫ا‬‫ﻋﺪد‬ ‫ﻟﻴﺲ‬ X
X≠4 ‫او‬ X+1=6
~ P(X)P(X)

16
(X-3)(X-4)=0‫و‬‫او‬
( 1 - 2 ) ‫تمرينات‬
� 1‫س‬
:‫اﻵﺗﻴﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﻣﻦ‬ ‫ﻣﻔﺘﻮﺣﺔ‬ ‫ﺟﻤﻠﺔ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫اﻛﺘﺐ‬
‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻠﺔ‬
N X<3 ( ‫أ‬
(‫ب‬
(‫ﺟـ‬
( ‫د‬
{10 , 8 , 6 , 4 , 2} 4 ‫ﻋﻠﻰ‬ ‫اﻟﻘﺴﻤﺔ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬ X (‫ﻫـ‬
( ‫و‬
‫ﺟﻤﻠﺘﻴﻦ‬ ‫ﻳﻤﺜﻞ‬ ‫اﻷزواج‬ ‫ﻫﺬه‬ ‫ﻣﻦ‬ ‫أي‬ ، ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫ﻤﻞ‬ ُ‫اﻟﺠ‬ ‫ﻣﻦ‬ ‫زوج‬ ‫ﻳﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫ﻳﻮﺟﺪ‬ � 2‫س‬
. Z ‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫اﻟﻌﻠﻢ‬ ‫ﻣﻊ‬ ‫ﻣﺘﻜﺎﻓﺌﺘﻴﻦ‬ ‫ﻣﻔﺘﻮﺣﺘﻴﻦ‬
X=-3 X=3 , X2
=9 (‫ﺟـ‬ X=2 , X2
=4 (‫ب‬ X-3=3 , 3X-5=X+7 ( ‫أ‬
(‫ﻫـ‬ ( ‫د‬
( ‫ز‬ 1 ‫ﻣﻦ‬ ‫أﺻﻐﺮ‬ ‫و‬ -1 ‫ﻣﻦ‬ ‫أﻛﺒﺮ‬ X ، X = 0 ( ‫و‬
‫ﻣﻊ‬ ‫اﻟﻤﻨﻔﻴﺔ‬ ‫ﻟﻠﺠﻤﻠﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ ‫اﻵﺗﻴﺔ‬ ‫ﻤﻞ‬ُ‫اﻟﺠ‬ ‫ﻣﻦ‬ ‫ﻣﻔﺘﻮﺣﺔ‬ ‫ﺟﻤﻠﺔ‬ ‫ﱠ‬‫ﻛﻞ‬ ِ‫ﻧﻒ‬ِ‫ا‬ � 3‫س‬
{ 1 , 2 , 3 , 4 , 5 } ‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫اﻟﻌﻠﻢ‬
X-1=4 X2
=16 (‫ﻫـ‬ X+2=4 X2
≠9 ( ‫د‬ ( ‫ﺟـ‬ X+4=7 (‫ب‬ 2X=4 ( ‫أ‬
{ 0 , 1 , 2 , ... , 9 } ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫ﻋﻨﺎﺻﺮ‬ X ، Y ‫أن‬ ‫ﻋﻠﻤﺖ‬ ‫إذا‬ � 4‫س‬
‫ﻣﺮﺗﺒﺔ‬ ‫أزواج‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﺠﻤﻞ‬ ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺎﻛﺘﺐ‬
X-Y=3 ( ‫أ‬
X+Y=15 (‫ب‬
3
5
Z (X-1)(X- ‫ــــــ‬ )(X-30)=0
Z X+5≥0
‫او‬
X+1=0 , (X+1)(2X+1)=0X2
-6X+5=0 , (X-1)(X-5)=0
(X-1)(X-2)=0 , 3 >X≥0
{10, 6 , 5 , 3} X2
-11X+30=0
N (X-1)(X-5)=0‫و‬ X>4

17
Quantifiered Propositions ‫اﻟﻤﺴﻮرة‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ [1�7�
: ً‫ا‬‫جزئي‬ ‫المسورة‬ ‫ات‬‫ر‬‫والعبا‬ ً‫ا‬‫كلي‬ ‫المسورة‬ ‫ات‬‫ر‬‫العبا‬ [1-7-1]
‫ﻋﻠﻴﻬﺎ‬ ‫ﻣﺘﻔﻖ‬ ‫ﺑﺮﻣﻮز‬ ‫اﻟﻜﻠﻤﺎت‬ ‫ﻋﻦ‬ ‫اﻻﺳﺘﻌﺎﺿﺔ‬ ً‫ﺎ‬‫ﻣﻤﻜﻨ‬ ‫ذﻟﻚ‬ ‫ﻳﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬ ‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ ‫ﻳﺤﺎول‬
: ‫ﻫﺎﻣﻴﻦ‬ ‫ﻣﻨﻄﻘﻴﻴﻦ‬ ‫رﻣﺰﻳﻦ‬ ‫ﻫﻨﺎ‬ ‫وﺳﻨﻘﺪم‬
‫»ﻣﻬﻤﺎ‬ : ‫ﻧﻘﻮل‬ ‫ﻓﺈﻧﻨﺎ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻳﺠﻌﻞ‬ A ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻣﻦ‬ ‫ﻋﻨﺼﺮ‬ ‫ﻛﻞ‬ ‫أن‬ ‫ﻧﺬﻛﺮ‬ ‫أن‬ ‫أردﻧﺎ‬ ‫إذا‬ : ً‫ﻻ‬‫او‬
«‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻓﺈن‬ A ‫ﻣﻦ‬ a ‫ﻛﺎن‬
«‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻳﻜﻮن‬ a∈ A ‫»ﻟﻜﻞ‬ ‫أو‬
:‫اﻵﺗﻲ‬ ‫اﻟﻨﺤﻮ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺨﺘﺰل‬ ‫رﻣﺰي‬ ‫ﺑﺸﻜﻞ‬ ‫اﻟﻘﻮل‬ ‫ﻫﺬا‬ ‫وﻳﻜﺘﺐ‬
. ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻓﺎن‬
‫اﻟﻌـﺒــﺎرة‬ ‫وﺗﺴﻤــﻰ‬ ‫اﻟﻜــﻠــﻲ‬ ‫اﻟـﻤﺴﻮر‬ ‫أو‬ (‫اﻟﺸﻤﻮل‬ ‫)دﻻﻟﺔ‬ ً‫ﺎ‬‫ﻛﻠﻴ‬ ً‫ا‬‫ر‬‫اﻟﺮﻣﺰ∀ﺳﻮ‬ ‫ﻳﺴﻤﻰ‬
. ً‫ﺎ‬‫ﻛﻠﻴ‬ ‫ﻣﺴﻮرة‬ ‫ﻋﺒﺎرة‬ ‫ﻓﺈن‬
: ً‫ﻼ‬‫ﻣﺜ‬
: ‫ﻳﺄﺗﻲ‬ ‫ﻛﻤﺎ‬ ‫ﻛﺘﺎﺑﺘﻬﺎ‬ ‫وﻳﻤﻜﻦ‬ X ‫ﻣﻜﺎن‬ ‫ﻳﻮﺿﻊ‬ ‫ﻃﺒﻴﻌﻲ‬ ‫ﻋﺪد‬ ‫ﻟﻜﻞ‬ ‫ﺻﺎﺋﺒﺔ‬
: ‫ﻧﻘﻮل‬ ‫ﻓﺈﻧﻨﺎ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﺗﺠﻌﻞ‬ A ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻋﻨﺎﺻﺮ‬ ‫ﺑﻌﺾ‬ ‫أن‬ ‫ﻧﺬﻛﺮ‬ ‫أن‬ ‫أردﻧﺎ‬ ‫إذا‬ : ً‫ﺎ‬‫ﺛﺎﻧﻴ‬
«‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﻳﺠﻌﻞ‬ A ‫ﻣﻦ‬ ‫ﻋﻨﺼﺮ‬ ‫اﻻﻗﻞ‬ ‫ﻓﻲ‬ ‫»ﻳﻮﺟﺪ‬
: ‫ﻛﺎﻵﺗﻲ‬ ‫رﻣﺰي‬ ‫ﺑﺸﻜﻞ‬ ‫اﻟﻜﻼم‬ ‫ﻫﺬا‬ ‫وﻧﻜﺘﺐ‬
(‫اﻟﻮﺟﻮد‬ ‫)دﻻﻟﺔ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫ﺑﺤﻴﺚ‬ ∃b∈A
‫أردﻧﺎ‬ ‫ﻓﺈذا‬ ً‫ﺎ‬‫ﺟﺰﺋﻴ‬ ‫ﻣﺴﻮرة‬ ‫ﻋﺒﺎرة‬ ، ∃b∈A ‫اﻟﻌﺒﺎرة‬ ‫وﺗﺴﻤﻰ‬ ً‫ﺎ‬‫ﺟﺰﺋﻴ‬ ً‫ا‬‫ر‬‫ﺳﻮ‬ ∃ ‫اﻟﺮﻣﺰ‬ ‫ﻳﺴﻤﻰ‬
: ‫ﻛﺘﺒﻨﺎ‬ Z ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ً ّ‫ﺣﻼ‬ X+1=2 ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫أن‬ ‫ﻧﻘﻮل‬ ‫أن‬ ً‫ﻼ‬‫ﻣﺜ‬
X+1=2 ‫ﺑﺤﻴﺚ‬ ∃X∈Z
: ‫ﺑﻘﻮﻟﻨﺎ‬ ‫ﺗﻘﺪم‬ ‫ﻣﺎ‬ ‫وﻧﺬﻛﺮ‬
.«‫ﻣﺤﻘﻘﺔ‬ X+1=2 ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺗﻜﻮن‬ ‫ﺑﺤﻴﺚ‬ X∈Z ‫ﻋﻨﺼﺮ‬ ‫اﻻﻗﻞ‬ ‫ﻓﻲ‬ ‫»ﻳﻮﺟﺪ‬
F(X)
F(a)
F(a)
∀ a∈AF(a)
∀ a∈AF(a)
(X+1)2
=X2
+2X+1
(X+1)2
=X2
+2X+1 ‫ﻓﺄن‬ ∀X∈N
G(x)
G(x)
G(b)
G(b)

18
: ‫المسورة‬ ‫ات‬‫ر‬‫العبا‬ ‫نفي‬ [1-7-2]
:‫اﻵﺗﻲ‬ ‫اﻟﻰ‬ ‫ﻧﻨﺘﺒﻪ‬ ‫اﻟﻤﺴﻮرة‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻧﻔﻲ‬ ‫ﻧﺮﻳﺪ‬ ‫ﻋﻨﺪﻣﺎ‬
: ‫اﻟﺼﻔﺘﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻓﻘﻂ‬ ‫وواﺣﺪة‬ ‫ﺑﻮاﺣﺪة‬ ‫ﺗﺘﺼﻒ‬ ‫أن‬ ‫ﻳﺠﺐ‬ ‫ﻋﺒﺎرة‬ ‫ﻛﻞ‬ ‫»إن‬
.«‫ﺧﺎﻃﺌﺔ‬ ‫أو‬ ‫ﺻﺎﺋﺒﺔ‬
: ‫اﻟﻌﺒﺎرة‬ ‫ﻧﻔﻲ‬ ً‫ﻼ‬‫ﻣﺜ‬ ‫أردﻧﺎ‬ ‫ﻓﻠﻮ‬ -
«‫ﻳﻨﺼﻔﻪ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻫﺬه‬ ‫ﻣﺮﻛﺰ‬ ‫ﻣﻦ‬ ‫ﻋﻠﻴﻪ‬ ‫اﻟﻨﺎزل‬ ‫اﻟﻌﻤﻮد‬ ‫ﻓﺈن‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ ‫اﻟﻤﺮﺳﻮم‬ ‫اﻟﻮﺗﺮ‬ ‫ﻳﻜﻦ‬ ‫»ﻣﻬﻤﺎ‬
: ‫ﻧﻘﻮل‬ ‫ﻓﺎﻧﻨﺎ‬
‫ﻻ‬ ‫ﻣﺮﻛﺰﻫﺎ‬ ‫ﻣﻦ‬ ‫ﻋﻠﻴﻪ‬ ‫اﻟﻨﺎزل‬ ‫اﻟﻌﻤﻮد‬ ‫أن‬ ‫ﺑﺤﻴﺚ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ ً‫ﺎ‬‫ﻣﺮﺳﻮﻣ‬ ‫واﺣﺪ‬ ‫وﺗﺮ‬ ‫اﻻﻗﻞ‬ ‫ﻓﻲ‬ ‫»ﻳﻮﺟﺪ‬
.« ‫ﻳﻨﺼﻔﻪ‬
: ‫اﻟﻘﻮل‬ ‫ﺧﻄﺄ‬ ‫إﺛﺒﺎت‬ ‫اردﻧﺎ‬ ‫-وإذا‬
:‫اﻟﻘﻮل‬ ‫ﺻﻮاب‬ ‫ﻧﺒﺮﻫﻦ‬ ‫أن‬ ‫ﻳﻜﻔﻲ‬ ‫ﻓﺈﻧﻪ‬ «6 ‫ﻋﻠﻰ‬ ‫اﻟﻘﺴﻤﺔ‬ ‫ﻳﻘﺒﻞ‬ 2 ‫ﻋﻠﻰ‬ ‫اﻟﻘﺴﻤﺔ‬ ‫ﻳﻘﺒﻞ‬ ‫ﻃﺒﻴﻌﻲ‬ ‫ﻋﺪد‬ ‫»ﻛﻞ‬
.«6 ‫ﻋﻠﻰ‬ ‫اﻟﻘﺴﻤﺔ‬ ‫ﻳﻘﺒﻞ‬ ‫وﻻ‬ 2 ‫ﻋﻠﻰ‬ ‫اﻟﻘﺴﻤﺔ‬ ‫ﻳﻘﺒﻞ‬ ‫واﺣﺪ‬ ‫ﻃﺒﻴﻌﻲ‬ ‫ﻋﺪد‬ ‫اﻻﻗﻞ‬ ‫ﻓﻲ‬ ‫»ﻳﻮﺟﺪ‬
:‫اﻟﻘﻮل‬ ‫ﻧﻔﻲ‬ ‫أردﻧﺎ‬ ‫وإذا‬ -
.«‫ﻓﻴﺜﺎﻏﻮرس‬ ‫ﻣﺒﺮﻫﻨﺔ‬ ‫ﻳﺤﻘﻖ‬ ‫ﻻ‬ ‫واﺣﺪ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ ‫اﻻﻗﻞ‬ ‫ﻓﻲ‬ ‫»ﻳﻮﺟﺪ‬
.«‫ﻓﻴﺜﺎﻏﻮرس‬ ‫ﻣﺒﺮﻫﻨﺔ‬ ‫ﻳﺤﻘﻖ‬ ‫ﻓﺈﻧﻪ‬ ‫اﻟﻘﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻳﻜﻦ‬ ‫»ﻣﻬﻤﺎ‬ ‫ﻗﻠﻨﺎ‬
:‫أن‬ ‫ﻗﺪﻣﻨﺎﻫﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻷﻣﺜﻠﺔ‬ ‫ﻣﻦ‬ ‫ﻳﻨﺘﺞ‬
~[ P(x) ‫ﻓﺄن‬ ∀x∈X ] ≡ ~ P(x) ‫ﻓﺄن‬ ∃x∈X
~[ P(x) ‫ﻓﺄن‬ ∃x∈X ] ≡ ~ P(x) ‫ﻓﺄن‬ ∀x∈X
� � 7 ‫ﻣﺜﺎل‬
:‫ﻳﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ً‫ﻼ‬‫ﻛ‬ ِ‫ﻧﻒ‬ِ‫ا‬
: ‫أن‬ ‫ﺣﻴﺚ‬ ‫ﻓﺈن‬ ∀X (1
X > 0 ‫ﻓﺈن‬ ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ X ‫ﻛﺎن‬ ‫إذا‬ :
: ‫أن‬ ‫ﺣﻴﺚ‬ ‫ﻓﺈن‬ ∃X (2
‫ﻣﻮﺟﺐ‬ ‫زوﺟﻲ‬ ‫ﻋﺪد‬ X :
(3
P(X)
P(X)
P(X)
P(X)
P ∨ [∃X∈R : X+3≥5 ]

19
≡ ~ ( ) ‫ﻓﺈن‬ ∃ (1
≤ ‫ﺣﻴﺚ‬ ‫ﻃﺒﻴﻌﻲ‬ ‫∃ﻋﺪد‬ : ~ P ( )
. ً‫ا‬‫ﺮ‬‫ﺻﻔ‬ ‫ﻳﺴﺎوي‬ ‫أو‬ ‫أﺻﻐﺮ‬ ‫ﻃﺒﻴﻌﻲ‬ ‫ﻋﺪد‬ ‫ﻳﻮﺟﺪ‬ : ‫وﺑﺎﻟﻜﻼم‬
≡~ (X) ‫ﻓﺈن‬ ∀ (2
‫ﻓﺈن‬ ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ا‬‫ﻋﺪد‬ X ‫ﻳﻜﻦ‬ ‫ﻣﻬﻤﺎ‬ : ‫وﺑﺎﻟﻜﻼم‬ ‫ﻣﻮﺟﺐ‬ ‫ﻏﻴﺮ‬ X ‫ﻓﺈن‬ ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ا‬‫∀ﻋﺪد‬ : ~P( )
. ‫ﻣﻮﺟﺐ‬ ‫ﻏﻴﺮ‬ X
( ) (3
: Tautology ‫الحاصل‬ ‫التحصيل‬ [1-7-3]
P ‫ﻓﺎن‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻟﻬﺬه‬ ‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻻﺣﺘﻤﺎﻻت‬ ‫ﺟﻤﻴﻊ‬ ‫وﻛﺎﻧﺖ‬ P ‫اﻟﻤﻨﻄﻘﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻟﺪﻳﻨﺎ‬ ‫ﻛﺎن‬ ‫اذا‬
. ً‫ﻼ‬‫ﺣﺎﺻ‬ ً‫ﻼ‬‫ﺗﺤﺼﻴ‬ ‫ﺗﺴﻤﻰ‬
� � 8 ‫ﻣﺜﺎل‬
‫؟‬ ً‫ﻼ‬‫ﺣﺎﺻ‬ ً‫ﻼ‬‫ﺗﺤﺼﻴ‬ ‫ﺗﺸﻜﻞ‬ P∨~ P ‫ﻫﻞ‬ ‫ﻋﺒﺎرة‬ P ‫ﻟﺘﻜﻦ‬
. ً‫ﻼ‬‫ﺣﺎﺻ‬ ً‫ﻼ‬‫ﺗﺤﺼﻴ‬ ‫ﺗﺸﻜﻞ‬ ∴
. ( Contradiction ) ‫تناقﺾ‬ ‫تدعى‬ ‫ﺧاﻃﺌة‬ ‫الصواب‬ ‫قيم‬ ‫جميع‬ ‫كان‬ ‫اذا‬ : ‫مالحظة‬
P ~ P P∨~ P
T
F
F
T
T
T
P X
XX
XX P X
P
XX
P X+3 < 5 :∀ X ∈ R
~ [ (X) ‫ﻓﺈن‬ ∀X ] X
~ [ P ( ) ‫ﻓﺎن‬ ∃ ]
~ ∧
X 0

20
(1-3) ‫تمرينات‬
� 1‫س‬
: ‫ﺑﺪﻟﻬﺎ‬ ً‫ﺎ‬‫ﺻﺤﻴﺤ‬ ‫ﻟﻴﺲ‬ ‫ﺳﺘﻌﻤﺎل‬ِ‫ا‬ ‫دون‬ ‫ﻣﻦ‬ ‫اﻵﺗﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻣﻦ‬ ‫ﻋﺒﺎرة‬ ‫ﻛﻞ‬ ‫ﻧﻒ‬ِ‫ا‬
.‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬ ‫اﻟﻤﺘﺸﺎﺑﻬﺔ‬ ‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﺟﻤﻴﻊ‬ ( ‫أ‬
. ‫ﻣﺘﻄﺎﺑﻘﺔ‬ ‫ﻏﻴﺮ‬ ‫اﻟﻤﺘﺸﺎﺑﻬﺔ‬ ‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﺑﻌﺾ‬ (‫ب‬
. ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻳﻜﻮن‬ ‫ﻓﺈﻧﻪ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻛﺎن‬ ‫إذا‬ (‫ﺟـ‬
. ‫ﺣﻞ‬ ‫ﻟﻬﺎ‬ ‫ﻟﻴﺲ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺑﻌﺾ‬ ( ‫د‬
. ‫ﻣﺴﺘﻄﻴﻞ‬ ‫رﺑﺎﻋﻲ‬ ‫ﺷﻜﻞ‬ ‫ﻛﻞ‬ (‫ﻫـ‬
( ‫و‬
(‫ح‬
� 2‫س‬
: ‫اﻵﺗﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﺧﻄﺄ‬ ‫أو‬ ‫ﺻﻮاب‬ ‫ﺑﻴﻦ‬
: ‫أن‬ ‫ﺣﻴﺚ‬ P ( ) ‫ﻓﺈن‬ ، ∀X ( ‫أ‬
X2
= X ‫ﻓﺈن‬ ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ X ‫ﻛﺎن‬ ‫إذا‬ : (X)
: ‫أن‬ ‫ﺣﻴﺚ‬ (X) ‫ﻓﺈن‬ ∃X (‫ب‬
X2
= X ، ‫ﻃﺒﻴﻌﻲ‬ ‫ﻋﺪد‬ : (X)
: ‫أن‬ ‫ﺣﻴﺚ‬ (X) ‫ﻓﺈن‬ ∀X (‫ﺟـ‬
. ‫ﻣﻮﺟﺐ‬ ‫ﻋﺪد‬ X2
‫ﻓﺈن‬ ً‫ﺎ‬‫ﺳﺎﻟﺒ‬ ً‫ا‬‫ﻋﺪد‬ X ‫ﻛﺎن‬ ‫إذا‬ : (X)
. ً‫ﻼ‬‫ﺣﺎﺻ‬ ً‫ﻼ‬‫ﺗﺤﺼﻴ‬ : ‫ﻣﻨﻄﻘﻴﺘﺎن‬ ‫ﻋﺒﺎرﺗﺎن‬ Q ، P ( ‫د‬
‫ﺗﻨﺎﻗﺾ‬ ~ P∧P : ‫ﻋﺒﺎرة‬ P (‫ﻫـ‬
. ً‫ﻼ‬‫ﺣﺎﺻ‬ ً‫ﻼ‬‫ﺗﺤﺼﻴ‬ ( P ↔ Q ) ↔ ( P ↔ Q ) : ‫ﻣﻨﻄﻘﻴﺘﺎن‬ ‫ﻋﺒﺎرﺗﺎن‬ P ، Q ( ‫و‬
( ∀X∈R : X < 8 ) ∧P
X
P
P
PX
P
P
Q ∧P → Q
Q: ∀X ∈ N : X2
= 25

212121
2 ‫ﻭﺍﻟﻤﺘﺒﺎﻳﻨﺎﺕ‬ ‫ﺍﻟﻤﻌﺎﺩﻻﺕ‬ : ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻔﺼﻞ‬
Y =|X| ‫الدالة‬ ‫ورسم‬ ‫المطلقة‬ ‫القيمة‬ [2-1]
‫مطلق‬ ‫على‬ ‫تحتوي‬ ‫التي‬ ‫المعادالت‬ ‫حل‬ [2-2]
‫بمتﻐيرين‬ ‫أنيتين‬ ‫معادلتين‬ ‫حل‬ [2-3]
‫ات‬‫ر‬‫الفت‬ [2-4]
‫واحد‬ ‫متﻐير‬ ‫في‬ ‫األولى‬ ‫الدرجة‬ ‫من‬ ( ‫اجحة‬‫ر‬‫المت‬ ) ‫المتباينة‬ ‫حل‬ [2-5]
‫واحد‬ ‫متﻐير‬ ‫في‬ ‫الثانية‬ ‫الدرجة‬ ‫من‬ ‫المتباينة‬ ‫حل‬ [2-6]
‫السلوكية‬ ‫االهداف‬
:‫االتية‬ ‫االهداف‬ ‫تحقيق‬ ‫الى‬ ‫الفصل‬ ‫هذا‬ ‫تدريس‬ ‫يهدف‬
‫المطلقة‬ ‫القيمة‬ ‫على‬ ‫يتعرف‬ -
‫مطلق‬ ‫على‬ ‫تحتوي‬ ‫معادلة‬ ‫يحل‬ -
‫بمتﻐيرين‬ ‫الثانية‬ ‫الدرجة‬ ‫من‬ ‫آنيتين‬ ‫معادلتين‬ ‫يحل‬ -
‫بمتﻐيرين‬ ‫االولى‬ ‫الدرجة‬ ‫من‬ ‫متباينة‬ ‫يحل‬ -
‫واحد‬ ‫بمتﻐير‬ ‫الثانية‬ ‫الدرجة‬ ‫من‬ ‫متباينة‬ ‫يحل‬ -

22
� Absolute Value ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ [ 2 � 1 �
( 2 - 15 ) ‫ﺗﻌﺮﻳﻒ‬
: ‫ﻳﺄﺗﻲ‬ ‫ﻛﻤﺎ‬ |X| ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻬﺎ‬ ‫ﻧﺮﻣﺰ‬ ‫واﻟﺘﻲ‬ X ‫اﻟﺤﻘﻴﻘﻲ‬ ‫ﻟﻠﻌﺪد‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻌﺮف‬ُ‫ﺗ‬
X,∀ X> 0
| | = , X = 0
-X , ∀ X < 0
� � 1 ‫ﻣﺜﺎل‬
: ‫ﻳﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻛﻞ‬ ‫ﻋﻦ‬ ‫اﻟﺤﻘﻴﻘﻲ‬ ‫ﻟﻠﻌﺪد‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺗﻌﺮﻳﻒ‬ ‫ﺑﺎﺳﺘﺨﺪام‬ ‫ﻋﺒﺮ‬
X ∈ R ‫ﺣﻴﺚ‬ | X-3 | ( ‫ب‬ ( ‫أ‬
X-3, ∀ > 3 (‫ب‬
| | = X = 3 ‫ﻻن‬ ( ‫أ‬
-X+ 3, ∀
: ‫اﻵﺗﻴﺔ‬ ‫ﺑﺎﻟﺨﻮاص‬ ‫ﺗﺘﻤﺘﻊ‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫أن‬ ( 2 - 15 ) ‫اﻟﺘﻌﺮﻳﻒ‬ ‫ﻣﻦ‬ ‫ﻳﻨﺘﺞ‬
(1
(2
(3
(4
(5
Y ≠ 0 ‫|ﺣﻴﺚ‬ ‫|ـــــــ‬ = ‫ــــــ‬
(6
(7
]
]
X
Y
|X|
|Y|
: ‫ﻣﻼﺣﻈﺔ‬
ً‫ﺎ‬‫ﻗﻴﻤ‬ X ، Y ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ِ‫اﻋﻂ‬
‫ﺻﺤﺔ‬ ‫ﻣـــــﻦ‬ ‫وﺗﺄﻛـــﺪ‬ ‫ﻋﺪدﻳﺔ‬
. ‫ﺑﻨﻔﺴﻚ‬ ‫اﻟﺨﻮاص‬
| X |≥0 ‫ﻓﺎن‬ ∀X∈R
| -X |= | X |‫∀ﻓﺎن‬X∈R
-| X |≤X≤ | X | ‫∀ﻓﺎن‬X∈R
| X . Y |=| X | . | Y | ‫∀ﻓﺎن‬X∈R
|X|2
= X2
,∀X∈R
| X + Y |≤| X | + | Y | ‫∀ﻓﺎن‬X,Y∈R
-a≤X≤a ‫ﻓﺎن‬ | X | ≤ a ‫ﻛﺎن‬ ‫اذا‬ ∀a 0 X∈R
x
>
0
X-3
X
0,
X<3
3 = 9 < 10
∴ 3− 10 = 10 − 3 >
3 = 9 < 10
∴ 3− 10 = 10 − 3 > 010 − 3( )> 0
3 = 9 < 10
∴ 3− 10 = 10 − 3 > 010 − 3( )> 0

23
� � 2 ‫ﻣﺜﺎل‬
Y=| X | ‫إرﺳﻢ‬
(2-15) ‫ﺗﻌﺮﻳﻒ‬ ‫ﺣﺴﺐ‬
، Y = X ‫اﻟﻤﺴﺘﻘﻴﻢ‬ : ً‫ﻻ‬‫او‬
: ‫اﻟﻤﺴﺘﻘﻴﻢ‬ : ً‫ﺎ‬‫ﺛﺎﻧﻴ‬
]
X Y ( X , Y )
0 0 ( 0, 0 )
1 1 ( 1 , 1 )
2 2 ( 2 , 2 )
X Y ( X , Y )
0 0 ( 0 , 0 ) ‫ﻓﺠﻮة‬
-1 1 ( -1 , 1 )
-2 2 ( -2 , 2 )
Y=| X |
( 1 , 1 )
( 2 , 2 )
Y = XY=-X
Y
( -2 , 2 )
(- 1, 1 )
( 0 , 0 )
X
X , X>0
Y= 0 , X=0
-X , X<0
X<0 , Y=-X
x ≥ 0

24
� � 3 ‫ﻣﺜﺎل‬
‫إرﺳﻢ‬
(2-15) ‫ﺗﻌﺮﻳﻒ‬ ‫ﺣﺴﺐ‬
‫اﻟﻤﺴﺘﻘﻴﻢ‬ : ً‫ﻻ‬‫او‬
‫اﻟﻤﺴﺘﻘﻴﻢ‬ : ً‫ﺎ‬‫ﺛﺎﻧﻴ‬: ً‫ﺎ‬‫ﺛﺎﻧﻴ‬
X Y ( Y , X )
1 3 ( 1 , 3 )
3 5 ( 3 , 5 )
X Y ( X, Y )
1 3 ( 1 , 3 ) ‫ﻓﺠﻮة‬
0 4 ( 0 , 4 )
Y
X
( 3,5 )
( 1,3 )
( 0,4)
Y=| X - 1 |+3
]
]
(X-1)+3 ,∀X≥1
(-X+1)+3 , ∀X<1
X+2 , ∀X≥1
-X+4 , ∀X<1
Y=
∴ Y =
∀X≥1 , Y=X+2
∀X<1 , Y=-X+ 4
Y=| X - 1 |+3

25
� ‫ﻣﻄﻠﻖ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺘﻮي‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ [ 2 � 2 �
� � 4 ‫ﻣﺜﺎل‬
. X ∈ R ‫ﺣﻴﺚ‬ : ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
:‫أن‬ ‫اﻟﺤﻘﻴﻘﻲ‬ ‫ﻟﻠﻌﺪد‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺗﻌﺮﻳﻒ‬ ‫ﻣﻦ‬ ‫ﻧﺴﺘﻨﺘﺞ‬
= |3X+6|
:‫اﻟﻨﻈﺎم‬ ‫ﺗﻜﺎﻓﻲء‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫﺬه‬ ‫إن‬
....... { X: X≥ -2 }‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ 3X+6=9
........... { X: X< -2 } ‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ X- 6 = 9
. ‫اﻟﺼﻔﺮ‬ ‫ﻳﺴﺎوي‬ ‫ﻓﻴﻬﺎ‬ y ‫ﻣﻌﺎﻣﻞ‬ ‫ﺣﻴﺚ‬ . x ،y ‫ﺑﺎﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﻣﻌﺎدﻟﺘﻴﻦ‬ ‫ﻧﻈﺎم‬ ‫اﻟﻨﻈﺎم‬ ‫ﻫﺬا‬ ‫ﱠ‬‫ﺪ‬ِ‫ﻌ‬ُ‫ﻧ‬ ‫ن‬َ‫أ‬ ‫ﻳﻤﻜﻨﻨﺎ‬
: ‫ﻫﻲ‬ ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﻫﺎﺗﻴﻦ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫إن‬
S1
= { 1 } ، S2
= { -5}
. S= S1
∪ S2
= { 1 , -5} ‫ﻫﻲ‬ ‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ∴
}
}
|3X+6|=9
-2≤X 0≤3X+6 3X+6 ‫اي‬ ‫ﻛﺎن‬ ‫إذا‬
-2>X 0>3X+6 -(3X+6)‫اي‬ ‫ﻛﺎن‬ ‫إذا‬
(2)
(1)
-3

26
� � 5 ‫ﻣﺜﺎل‬
. : ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
: ‫اﻟﻨﻈﺎم‬ ‫ﺗﻜﺎﻓﻰء‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻓﺎن‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺗﻌﺮﻳﻒ‬ ‫ﻣﻦ‬
✽X3
- 8 = 0 , ∀X ≥0 ⇒ X3
= 8 ⇒ X = 2
S1
= { 2}
✽- X3
- 8 = 0 , ∀X <0 ⇒ X3
= - 8 ⇒ X = -2
S2
= { -2}
. S = S1
∪ S2
= { 2 , -2}
� � 6 ‫ﻣﺜﺎل‬
. : ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
: ‫اﻟﻨﻈﺎم‬ ‫ﺗﻜﺎﻓﻰء‬ X2
+ |X| - 12 = 0 ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻓﺎن‬ ‫اﻟﻤﻄﻠﻘﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺗﻌﺮﻳﻒ‬ ‫ﻣﻦ‬
✽X2
+ X - 12 = 0 ,∀X ≥0⇒ ( X + 4 )( X - 3 ) = 0
X = 3 ‫او‬ ‫؟‬ ‫ﻟﻤﺎذا‬ ‫ﻳﻬﻤﻞ‬ X = -4 ‫اﻣﺎ‬
S1
= { 3} ∴
✽X2
- X - 12 = 0 ,∀X <0⇒ ( X - 4 )( X + 3 ) = 0
X = -3 ‫او‬ ‫؟‬ ‫ﻟﻤﺎذا‬ ‫ﻳﻬﻤﻞ‬ X = 4 ‫اﻣﺎ‬
S2
= { -3} ∴
S = S1
∪S2
= { 3 , -3}
∀X ∈ R ، X2
|X| - 8 = 0
X2
|X| - 8 = 0
∀X ∈ R ، X2
+ |X| - 12 = 0

27
�‫ﺑﻤﺘﻐﻴﺮﻳﻦ‬ ‫آﻧﻴﺘﻴﻦ‬ ‫ﻣﻌﺎدﻟﺘﻴﻦ‬ ‫ﺣﻞ‬ [ 2 � 3 �
‫وﺣﻴﻨﺬاك‬ ،ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ‫ﺑﻤﺘﻐﻴﺮﻳﻦ‬ ‫اﻷوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫ﻣﻌﺎدﻟﺘﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻣﺆﻟﻒ‬ ‫ﻧﻈﺎم‬ ‫ﺣﻞ‬ ‫اﻟﻄﺎﻟﺐ‬ ‫ﺗﻌﻠﻢ‬ ‫ﻟﻘﺪ‬
‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺈن‬ ، ‫اﻟﺜﺎﻧﻴﺔ‬ ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ً‫ﻼ‬‫ﺣ‬ S2
، ‫اﻷوﻟﻰ‬ ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ً‫ﻼ‬‫ﺣ‬ S1
‫ﻛﺎن‬ ‫اذا‬ ‫اﻵﺗﻲ‬ ‫وﺿﺤﻨﺎ‬
‫و‬ ‫اﻟﺮﺑﻂ‬ ‫ﺑﺎداة‬ ‫ﻣﺮﺑﻮﻃﺘﻴﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ . S = S1
∩ S2
.S = S1
∪S2
‫ﻫﻮ‬ ‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ّ‫ﻓﺈن‬ ‫أو‬ ‫اﺑﻂ‬‫ﺮ‬‫اﻟ‬ ‫ﻛﺎن‬ ‫إذا‬ ‫أﻣﺎ‬
� � 7 ‫ﻣﺜﺎل‬
: ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺠﺪ‬ Y ، X ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬
ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ‫و‬ ً‫ﺎ‬‫ﺗﺤﻠﻴﻠﻴ‬
X - 2Y = 5 . . . . . (1)
2X + Y = 0 . . . . . (2)
: 2 ‫ﺑﺎﻟﻌﺪد‬ (2) ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻀﺮب‬ : ً‫ﺎ‬‫ﺗﺤﻠﻴﻠﻴ‬
X - 2Y = 5 . . . . . (1)
4X + 2Y = 0 . . . . . (2)
:(1) ‫ﻓﻲ‬ ‫ﻧﻌﻮض‬
. ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺗﻤﺜﻞ‬ ‫وﻫﻲ‬ .{( 1 ,- 2 )} = ‫ﻣﺞ‬ ∴
R
‫ﺑﺎﻟﺠﻤﻊ‬
5X=5 ⇒ X = 1
⇒ Y = -2
1 - 2Y = 5

28
X - 2Y = 5 : ‫اﻟﻤﺴﺘﻘﻴﻢ‬ : ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬
2X + Y = 0 : ‫اﻟﻤﺴﺘﻘﻴﻢ‬
� � 8 ‫ﻣﺜﺎل‬
‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺠﺪ‬ x.y ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ R ‫ﻛﺎﻧﺖ‬ ‫اذا‬
‫ﻓﻴﻜﻮن‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﺴﺐ‬ ‫ﺑﺎﻟﻌﻜﺲ‬ ‫او‬ y ‫ﺑﺪﻻﻟﺔ‬ x ‫ﻧﺠﺪ‬ ‫ﺣﻴﺚ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﺑﻄﺮﻳﻘﺔ‬ ‫ﻳﺤﻞ‬ ‫اﻟﻨﻈﺎم‬ ‫ﻫﺬا‬
‫ﻓﻴﻜﻮن‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺗﻌﻮض‬ X= 1 + Y
2 ‫ﻋﻠﻰ‬ ‫ﺑﺎﻟﻘﺴﻤﺔ‬
: ‫اﻟﻤﺴﺘﻘﻴﻢ‬
X Y ( X , Y )
0 -5/2 ( 0 , -5/2 )
1 -2 ( 1 , -2 )
5 0 ( 5 , 0 )
: ‫اﻟﻤﺴﺘﻘﻴﻢ‬
X, YYX
(0,0 )00
(1 ,-2)-21
(-1 ,2)2-1
X - Y = 1
X2
+ Y2
= 13
L2
L1
(5,0)
(1,-2)
(-1,2) L1
L2
(1+y)2
+ y2
= 13 ⇒ 2 y2
+ 2y-12 = 0
y2
+ y-6 = 0 ⇒ (y+3)(y-2)=0
y + 3 = 0 ⇒ y = -3 ⇒ x = -2 ⇒ (-2,3)
y - 2 = 0 ⇒ y = 2 ⇒ x = 3 ⇒ (3,2)
∴S= (-2,-3),(3,2){ }

29
� � 9 ‫ﻣﺜﺎل‬
‫اﻟﺤﺬف‬ ‫ﺑﻄﺮﻳﻘﺔ‬ x,y ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ R ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫اﻵﺗﻲ‬ ‫اﻟﻤﺜﺎل‬ ‫ﺣﻞ‬
2x2
− 3y2
= -46 , x2
+ y2
= 17
‫)ﻳﺘﺮك‬ ‫واﻟﺘﻌﻮﻳﺾ‬ ‫اﻟﺤﺬف‬ ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ ‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻳﻤﻜﻦ‬ ‫اذن‬ ،‫اﻟﺪرﺟﺔ‬ ‫ﻧﻔﺲ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ان‬ ‫ﺑﻤﺎ‬
(‫ﻟﻠﻄﺎﻟﺐ‬

x2
+ y2
= 17....1
2x2
− 3y2
= -46....2
---------------------------
3x2
+ 3y2
= 51
2x2
− 3y2
= -46
----------------------
5x2
= 5 ⇒ x2
= 1⇒ x= m1
x=1 ⇒ (1)2
+ y2
= 17 ⇒ y2
= 16 ⇒ y = m4 ⇒ (1,4),(1,−4)
x=-1 ⇒ (-1)2
+ y2
= 17 ⇒ y2
= 16 ⇒ y = m4 ⇒ (−1,4),(−1,−4)
S = (1,4),(1,−4),(−1,4),(−1,−4){ }
�3 ‫ﻓﻲ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫)ﺑﻀﺮب‬
‫اﻟﺨﻼﺻﺔ‬
‫ﺑﻄﺮﻳﻘﺘﻲ‬ ‫ﻓﺘﺤﻞ‬ (‫اﻟﺜﺎﻧﻴﺔ‬ ‫او‬ ‫)اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻧﻔﺲ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ (1
‫اﻟﺘﻌﻮﻳﺾ‬ * ‫اﻟﺤﺬف‬ *
‫ﺑﻄﺮﻳﻘﺔ‬ ‫ﻓﺘﺤﻞ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫واﻻﺧﺮى‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫اﺣﺪﻫﻤﺎ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ (2
‫اﻟﺘﻌﻮﻳﺾ‬

30
� Intervals ‫ات‬‫ﺮ‬‫اﻟﻔﺘ‬ [ 2 � 4 �
a ، b ∈ R ، a < b ‫ﻟﻴﻜﻦ‬
: ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺗﺴﻤﻰ‬ ( 1
‫وﻧــﺮﻣــﺰ‬ b ‫اﻟـــﻰ‬ a ‫ﻣــﻦ‬ Closed Interval ‫اﻟﻤﻐﻠﻘﺔ‬ ‫اﻟﻔﺘﺮة‬ { X : X ∈ R ، a ≤ X ≤ b }
‫ﻟﻨﻘﻄﺔ‬ ‫رﻣﺰﻧﺎ‬ ‫ﺣﻴﺚ‬ ( 2 - 1 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ ‫وﺗﻤﺜﻞ‬ [ a , b ] ‫ﺑﺎﻟـــﺮﻣﺰ‬ ‫ﻟﻬـــﺎ‬
‫ﻟﻬــﺬه‬ ‫اﻟﻨﻬـــﺎﻳﺔ‬ ‫وﻟﻨﻘﻄﺔ‬ ( a ) ‫ﺑﺎﺣﺪاﺛﻴﻬﺎ‬ ‫اﻟﻤﻐﻠﻘﺔ‬ ‫اﻟﻔﺘﺮة‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫ﻟﻠﻘﻄﻌﺔ‬ ‫اﻟﺒﺪاﻳﺔ‬
‫ﺑﻴﻦ‬ ‫ﺗﻘﺎﺑﻞ‬ ‫وﺟﻮد‬ ‫ﻳﻼﺣﻆ‬ ( ‫و‬ ) ‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ذﻛﺮ‬ ‫اﻟﺸﻜﻞ‬ ‫ﻫﺬا‬ ‫ﻋﻠﻰ‬ ‫أﻫﻤﻠﻨﺎ‬ ‫ﻟﻘﺪ‬ ( b ) ‫ﺑﺎﺣﺪاﺛﻴﻬﺎ‬ ‫اﻟﻘﻄﻌﺔ‬
a b ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻧﻘﺎط‬ ‫وﻣﺠﻤﻮﻋﺔ‬ [a , b ] ‫اﻟﻔﺘﺮة‬ ‫إﻟﻰ‬ ‫اﻟﻤﻨﺘﻤﻴﺔ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬
a b
( 2 - 1 ) ‫اﻟﺸﻜﻞ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻧﺴﻤﻲ‬ ( 2
‫ﻋﻠﻰ‬ ‫وﺗﻤﺜﻞ‬ ( b ) ‫اﻟـﻰ‬ ( a ) ‫ﻣﻦ‬ Open Interval ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫اﻟﻔﺘﺮة‬
( 2 - 2 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬
a b
( 2 - 2 ) ‫اﻟﺸﻜﻞ‬
‫ﻓﻲ‬ a , b ‫اﻟﻌﺪدﻳﻦ‬ ‫ﺣﻮل‬ ‫واﻟﺪاﺋﺮﺗﻴﻦ‬ ‫أن‬ ‫اﻟﺤﺎﻟﺔ‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ ‫ﻳﻼﺣﻆ‬
. ‫ذﻟﻚ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺪﻻن‬ ‫اﻟﺸﻜﻞ‬
b ∉ (a , b) , a ∉( a , b)
(a,b) ={X:X∈R,a<X<b}

31
: ‫ﻣﻦ‬ ِ‫ﻼ‬‫ﻛ‬ ‫ﻧﺴﻤﻲ‬ ( 3
( a , b ] = { X : X ∈ R ، a < X ≤ b }
[ a , b ) = { X : X ∈ R ، a ≤ X < b }
‫اﻷول‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺗﻤﺜﻞ‬ a < b ‫ﺣﻴﺚ‬ ( Half Open ‫اﻟﻤﻔﺘﻮﺣﺔ‬ ‫ﻧﺼﻒ‬ ‫)أو‬ ‫اﻟﻤﻐﻠﻘﺔ‬ ‫ﻧﺼﻒ‬ ‫اﻟﻔﺘﺮة‬
( 2 - 3 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬
a b
( 2 - 3 ) ‫اﻟﺸﻜﻞ‬
( 2 - 4 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺗﻤﺜﻞ‬
a b
( 2 - 4 ) ‫اﻟﺸﻜﻞ‬
: ‫ﻫﻲ‬ ‫ﺗﺴﺎوﻳﻪ‬ ‫أو‬ ( a ) ‫اﻟﺤﻘﻴﻘﻲ‬ ‫اﻟﻌﺪد‬ ‫ﻋﻠﻰ‬ ‫ﺗﺰﻳﺪ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ( 4
( 2 - 5 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫وﺗﻤﺜﻠﻬﺎ‬ { X : X ∈ R ، X ≥ a }
( 2 - 6 ) ‫اﻟﺸﻜﻞ‬ ‫ﻳﻤﺜﻠﻬﺎ‬ { X : X∈R ، X > a } ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫ﻛﻤﺎ‬
a a
( 2 - 6 ) ‫اﻟﺸﻜﻞ‬ ( 2 - 5 ) ‫اﻟﺸﻜﻞ‬
( a ) ‫اﻟﺤﻘﻴﻘﻲ‬ ‫اﻟﻌﺪد‬ ‫ﺗﺴﺎوي‬ ‫اﻟﺘﻲ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ( 5
‫ﻫﻲ‬ ‫ﺗﺼﻐﺮه‬ ‫أو‬
{ X : X ∈R ، X ≤a }
( 2 - 7 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻴﻤﺜﻠﻬﺎ‬
{ X : X ∈R ، X < a } ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻣﺎ‬
( 2 - 8 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻴﻤﺜﻠﻬﺎ‬
a a
( 2 - 8 ) ‫اﻟﺸﻜﻞ‬ ( 2 - 7 ) ‫اﻟﺸﻜﻞ‬
(5) ‫و‬ (4) ‫ﻓﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬
‫ﻏﻴﺮ‬ ‫ﻋﺪدﻳﺔ‬ ‫ﻣﺠﻤﻮﻋﺎت‬ ‫ﺗﺪﻋﻰ‬
(‫)ﺷﻌﺎع‬ ‫ﻣﺤﺪدة‬

32
� � 1 ‫ﻣﺜﺎل‬
‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺜﻞ‬ X = [ 1 , 6 ] ، Y = [ 3 , 8 ]‫ﻟﺘﻜﻦ‬
1 ) X∩Y 2 ) X∪Y
3 ) X - Y 4 ) Y - X
‫ﻓﺘﺮة‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻨﺎﺗﺞ‬ ‫اﻛﺘﺐ‬ ‫ﺛﻢ‬
1 3 6 8
1 ) X ∩ Y = [ 3 , 6 ] 3 ) X - Y = [ 1 , 3 )
2 ) X ∪ Y = [ 1 , 8 ] 4 ) Y - X = ( 6 , 8 ]
� � 2 ‫ﻣﺜﺎل‬
‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ { X : X ≥ -3 } ∪ ( -5 , 2 ] ( 1 ‫ﻣﺜﻞ‬
‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ { X : X ≥ -3 } ∩ ( -5 ,2] ( 2
-5 -3 0 2
∴ 1 ) { X : X ≥ -3 } ∪ ( -5 , 2 ] = { X : X >-5 }
2 ) { X : X ≥ -3 } ∩ ( -5 , 2 ] = [-3 , 2 ]

33
� ‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ (‫اﺟﺤﺔ‬‫ﺮ‬‫)اﻟﻤﺘ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ [ 2 � 5 �
g(X) ، f(X )‫ﺣﻴﺚ‬ g(X) < f(X ): ‫ﺑﺎﻟﺸﻜﻞ‬ ‫ﺗﻜﺘﺐ‬ ‫واﻟﺘﻲ‬ ( X ) ‫ﻣﺘﻐﻴﺮ‬ ‫ﺗﺤﻮي‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫إن‬
. ( X ) ‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ Inequality ‫ﻣﺘﺒﺎﻳﻨﺔ‬ ‫ﺗﺴﻤﻰ‬ ‫ﻣﻔﺘﻮﺣﺎن‬ ‫ﺟﻤﻠﺘﺎن‬
‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ ( X ) ‫ﻟـ‬ ‫أﻋﻄﻴﺖ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﺳﺘﻚ‬‫ر‬‫د‬ ‫ﻣﻦ‬ ‫ﺗﻌﻠﻢ‬ ‫وﻛﻤﺎ‬
‫ﻛﻤﺎ‬ ‫اﻟﻤﺘﻜﺎﻓﺌﺔ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺎت‬ ‫ﻌﺮف‬ُ‫ﺗ‬‫و‬ . ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻫﺬه‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أوﺟﺪﻧﺎ‬ ‫ﻧﻘﻮل‬ ، ‫ﺻﺎﺋﺒﺔ‬ ‫ﻋﺒﺎرة‬ ‫وﺟﻌﻠﻬﺎ‬
. ‫اﻟﻤﺘﻜﺎﻓﺌﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺮﻓﺖ‬ُ‫ﻋ‬
( 2 - 16 ) ‫ﺗﻌﺮﻳﻒ‬
‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻬﻤﺎ‬ ‫ﻛﺎن‬ ‫إذا‬ ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻣﻜﺎﻓﺌﺔ‬ ‫ﻣﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻋﻦ‬ ‫ﻧﻘﻮل‬
.‫ﻧﻔﺴﻬﺎ‬ ‫اﻟﺤﻞ‬
. ‫اﻟﺤﺪود‬ ‫ﻛﺜﻴﺮة‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻓﻴﻬﺎ‬ ‫ﻳﻜﻮن‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺎت‬ ‫ﺑﺤﻞ‬ ‫اﻟﺒﻨﺪ‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫ﺳﻨﻬﺘﻢ‬
� � 1 ‫ﻣﺜﺎل‬
3X +1< X +5 : ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
. ‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫وﺿﻊ‬ ، R ‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬
3X +1< X + 5
‫اﻟﺤﻘﻞ‬ ‫ﺧﻮاص‬ 2X +1 < 5 ⇐
⇐
. ‫اﻟﺤﻘﻞ‬ ‫ﺧﻮاص‬ 2X < 4 ⇐
f(X) < g(X)h(X) < I(X)
g(X) ، f(X)
2X + 1 + (-1) < 5 +(-1)
3X + 1+(-X) <X + 5 + (-X)

34
(2X) ‫< ـــــ‬ 4 ‫ــــــ‬ ⇐
. ‫اﻟﺤﻘﻞ‬ ‫ﺧﻮاص‬ X < 2 ⇐
{ X : X ∈R ، X < 2 } = ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ∴
2
‫ﻣﻦ‬ ‫ﻣﺘﺒﺎﻳﻨﺘﻴﻦ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺆﻟﻒ‬ ‫اﻟﻨﻈﺎم‬ ‫ﻫﺬا‬ ‫ﺗﺤﻘﻖ‬ ‫اﻟﺘﻲ‬ X ‫ﻗﻴﻤﺔ‬ ‫ﻓﺈن‬ ‫و‬ ‫اﺑﻂ‬‫ﺮ‬‫ﺑﺎﻟ‬ ‫ﻣﺘﺒﺎﻳﻨﺘﻴﻦ‬ ‫رﺑﻄﻨﺎ‬ ‫إذا‬
S2
‫وإﻟﻰ‬ ‫اﻷوﻟﻰ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ S1
‫إﻟﻰ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫أن‬ ‫ﻳﺠﺐ‬ ‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ ‫اﻷوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬
:‫ﻳﻌﻨﻲ‬ ‫وﻫﺬا‬ : S1
∩ S2
‫إﻟﻰ‬ ‫أي‬ . ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
: ‫ﻫﻲ‬ ‫و‬ ‫اﺑﻂ‬‫ﺮ‬‫واﻟ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺘﻴﻦ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻜﻮن‬ ‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ّ‫أن‬
‫ﻫﻲ‬ ‫أو‬ ‫اﺑﻂ‬‫ﺮ‬‫واﻟ‬ ‫ﻣﺘﺒﺎﻳﻨﺘﻴﻦ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻜﻮن‬ ‫اﻟﻨﻈﺎم‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫ﻣﺸﺎﺑﻪ‬ ‫ﺑﺸﻜﻞ‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫أن‬ ‫وﻳﻤﻜﻨﻨﺎ‬
� � 2 ‫ﻣﺜﺎل‬
: ‫ﻟﻠﻨﻈﺎم‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬ ( R ) ‫ﻫﻲ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬
. ‫اﻷﻋﺪاد‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ ‫إﺟﺎﺑﺘﻚ‬ ‫ﻞ‬ّ‫ﺜ‬‫ﻣ‬ 2X + 3 < 6 ‫و‬ 5X + 11 < 1
= { X : X <-2 }‫ﻫﻲ‬ ‫اﻷوﻟﻰ‬ ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
= { } ‫ﻫﻲ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
S : ‫ﻫﻲ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺘﻴﻦ‬ ‫ﻟﻨﻈﺎم‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
1
2
1
2
S = S1
∪ S2
S1
S2 X:X < ‫ ـــــ‬3
2
S = S1
∩S2
( )( )

35
3
2
3
2
3
2
3
2
3
2
{ X : X < -2 } ∩ { X : X < ‫ــــــ‬ }
S1
∩S2
=S1
=
S = S1
∩S2
=
-2 ‫ــــــ‬
‫ﻧﻔﺴﻬﺎ‬ S1
‫ﻫﻲ‬ S1
، S2
‫ﺑﻴﻦ‬ ‫اﻟﻤﺸﺘﺮﻛﺔ‬ ‫اﻟﻌﻨﺎﺻﺮ‬
{ X : X < -2 ، X ∈R }
� � 3 ‫ﻣﺜﺎل‬
: ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﻤﺜﺎل‬ ‫ﻓﻲ‬ ‫أو‬ ‫اﺑﻂ‬‫ﺮ‬‫ﺑﺎﻟ‬ ‫و‬ ‫اﺑﻂ‬‫ﺮ‬‫اﻟ‬ ‫ﻋﻮض‬
2X + 3 < 6 ‫أو‬ 5x + 11 < 1 : ‫ﻟﻠﻨﻈﺎم‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
-2 ‫ــــــ‬
S2
‫ﻫﻲ‬ ً‫ﺎ‬‫ﻣﻌ‬ ‫ﻛﻠﻴﻬﻤﺎ‬ ‫ﻓﻲ‬ ‫أو‬ S2
‫أو‬ S1
‫ﻓﻲ‬ ‫اﻟﻤﻮﺟﻮدة‬ ‫اﻟﻌﻨﺎﺻﺮ‬ ‫أن‬ ‫ﻧﻼﺣﻆ‬
S= { X : < -2 ‫ـــــ‬ >X }3
2
‫و‬
S2
∪ S1
= {X: X <‫ـــــــ‬ X < -2}‫أو‬
S = {X:X ∈R , X< ‫ـــــــ‬}

36
� � 4 ‫ﻣﺜﺎل‬
|X-2| > 5 ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬ ‫اﻟﺘﻌﻮﻳﺾ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻮ‬ ( R ) ‫ﻛﺎن‬ ‫إذا‬
X -2 , ∀ X≥ 2
= ‫أو‬
2- X , ∀ X <2
}X-2 -
2-X > 5 ‫أو‬ X-2 > 5 ⇔ |X-2| > 5 ∴
: ‫ﻫﻲ‬ ‫اﻟﻤﻄﻠﻮﺑﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أن‬ ‫ﻧﺠﺪ‬ ‫اﻟﻨﻈﺎم‬ ‫ﻫﺬا‬ ‫وﺑﺤﻞ‬
S1
∪ S2 = { X : X ∈ R ، X > 7 } ∪ { X : X ∈R ، X < -3 }
S2
-3 7 S1
� � 5 ‫ﻣﺜﺎل‬
x ∈ R ‫ﺣﻴﺚ‬ x+1 ≤ 2 ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬
52‫ص‬ (7) ‫ﺧﺎﺻﻴﺔ‬ ‫ﺣﺴﺐ‬ ‫ﻣﺒﺎﺷﺮة‬ ‫ﺣﻠﻬﺎ‬ ‫ﻳﻤﻜﻦ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻫﺬه‬ ‫ان‬ ‫ﻻﺣﻆ‬
‫ﻓﻴﻜﻮن‬
‫ﻳﻨﺘﺞ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﺪود‬ ‫اﻟﻰ‬ (-1) ‫ﺑﺎﺿﺎﻓﺔ‬
-2+ -1( ) ≤ x+1+ −1( ) ≤ 2+ −1( )
−3 ≤ x ≤ 1
∴s = −3,1[ ]
x+1 ≤ 2 ⇒ -2 ≤ x+1 ≤ 2

37
‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫ﻣﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ [ 2 � 6 �
‫مبرهنة‬
: ‫ﻓﺈن‬ ً‫ﺎ‬‫ﻣﻮﺟﺒ‬ ً‫ﺎ‬‫ﺣﻘﻴﻘﻴ‬ ً‫ا‬‫ﻋﺪد‬ (a ) ‫ﻛﺎن‬ ‫إذا‬
[- a , a ] ‫اﻟﻔﺘﺮة‬ ‫ﻫﻲ‬ X2
≤ a2
‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ (1
(- a , a ) ‫اﻟﻔﺘﺮة‬ ‫ﻫﻲ‬ X2
< a 2
‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ (2
: (2 ‫اﻟﺒﺮﻫﺎن‬
(b <0 ‫و‬ a>0 ) ‫اﻣﺎ‬
‫ﻓـــﺎن‬ ‫ﺻﻔﺮ‬ > a . b ‫ﻛﺎن‬ ‫اذا‬ : ‫اﻟﻘﺎﻋﺪة‬ ‫ﻧﺴﺘﻨﺞ‬ ‫اﻟﻤﺮﺗﺐ‬ ‫اﻟﺤﻘﻞ‬ ‫ﺧﻮاص‬ ‫وﻣﻦ‬
(b >0 ‫و‬ a <0 ) ‫أو‬
[(X - a ) < 0 ‫و‬ (X +a ) > 0] ‫أو‬ [(X - a ) > 0 ‫و‬ (X+ a ) <0]
(- a , a) ∪ ϕ = (-a , a)
. ‫ﻟﻠﻄﺎﻟﺐ‬ ‫ﻧﺘﺮﻛﻬﺎ‬ ‫واﻟﺘﻲ‬ (1) ‫ﺑﺮﻫﻨﺔ‬ ‫ﻳﻤﻜﻦ‬ ‫ﻣﻤﺎﺛﻠﺔ‬ ‫وﺑﻄﺮﻳﻘﺔ‬
� � 6 ‫ﻣﺜﺎل‬
: ‫ﻫﻲ‬ ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺈن‬ X 2
< 9 ‫ﻛﺎن‬ ‫إذا‬
[ - 3, 3] ‫ﻫﻲ‬ ‫ﻟﻠﻤﺘﺒﺎﻳﻨﺔ‬ ‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﺈن‬ X2
≤ 9 ‫ﻛﺎن‬ ‫واذا‬ . (-3,3)
/ X2
≤ 9 ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﻬﻲ‬ X2
> 9 ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أﻣﺎ‬
R / [-3 ,3 ] ‫أي‬
/ X2
<9 ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻫﻲ‬ X2
≥9 ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻠﻮل‬ ‫وﻣﺠﻤﻮﻋﺔ‬
SR / (-3 ,3 ) ‫أي‬
}
( X - a ) ( X + a ) <0 ⇐ X2
- a2
<0 ⇐ X2
< a2
[X < a ‫و‬ X>-a ] ‫أو‬ [X > a ‫و‬ X < - a ]
⇒
⇒
R
R

38
� � 7 ‫ﻣﺜﺎل‬
≥ 5 : ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
: ‫اﻟﻨﻈﺎم‬ ‫ﺗﻜﺎﻓﺊ‬ |2X+5| ≥ 5 ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫إن‬
[ - (2X+5) ≥ 5 ] ‫أو‬ [ 2X + 5≥5 ]
[ ] ‫أو‬ [2 2 ≥0] ⇐
⇐
‫اﻟﺤﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
7 >|2X+5|
2X +5 , ∀ X≥ ‫ـــ‬ ‫ــ‬
2X +5 =
-( 2X+ 5) , ∀ X < ‫ـــــــــ‬
} - 5
2
- 5
2
7 >
7 >
7 >
X12 > - 2X ≥ 10
[ -6 <X ≤-5] [ 1>X ≥0]‫أو‬
( -6,-5] ∪[ 0,1)=
>
:‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﻟﺤﻞ‬
‫وﺟﺪ‬ ‫ان‬ ‫اﻟﻤﻄﻠﻖ‬ ‫ﻧﻌﺮف‬ *
:‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻻﻋﺪاد‬ ‫ﺣﻘﻞ‬ ‫ﺧﻮاص‬ ‫ﻧﺴﺘﺨﺪم‬ *
(0) ‫اﻟﺠﻤﻊ‬ ‫ﻋﻤﻠﻴﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﺤﺎﻳﺪ‬ ‫اﻟﻌﻨﺼﺮ‬ ‫اﻟﺘﺠﻤﻴﻊ‬ ‫ﺧﺎﺻﻴﺔ‬ ‫اﻟﺠﻤﻌﻲ‬ ‫اﻟﻨﻈﻴﺮ‬ ‫)اﺿﺎﻓﺔ‬
‫ﻋﻤﻠﻴﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﺤﺎﻳﺪ‬ ‫اﻟﻌﻨﺼﺮ‬ ‫اﻟﺘﺠﻤﻴﻊ‬ ‫ﺧﺎﺻﻴﺔ‬ ‫اﻟﻀﺮﺑﻲ‬ ‫اﻟﻨﻈﻴﺮ‬ ‫ﻓﻲ‬ ‫اﻟﻀﺮب‬
((1) ‫اﻟﻀﺮب‬ ‫ﻋﻤﻠﻴﺔ‬
R ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻻﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺿﻤﻦ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺔ‬ ‫ﺣﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻣﻦ‬ ‫اﻟﺴﻠﺴﻠﺔ‬ ‫ﻫﺬه‬ ‫ﺑﻌﺪ‬ *

39
/ 1‫س‬
‫ﻛﺎن‬ ‫اذا‬
A∪ B ، A∩B ، A - B، B - A ‫ﺟﺪ‬
/ 2‫س‬
(‫ب‬ Y =| X + 2 | - 5 ‫اﻟﺪاﻟﺔ‬ ‫(ارﺳﻢ‬ ‫أ‬
/ 3‫س‬
: ‫اﻟﺤﻞ‬ ‫ﻣﻦ‬ ‫ﺗﺤﻘﻖ‬ ‫ﺛﻢ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
(‫ﻫـ‬ ( ‫أ‬
(‫و‬ (‫ب‬
X2
-2 |X| - 15= 0 (‫ﺟـ‬
(‫د‬
/ 4‫س‬
:‫اﻵﺗﻴﺘﻴﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﺣﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
( ً‫ﺎ‬‫)ﺑﻴﺎﻧﻴ‬ 2X + Y = 4 ، X-Y = -1 ( ‫أ‬
(ً‫ﺎ‬‫)ﺗﺤﻠﻴﻠﻴ‬ 4X + 3Y= 17 ، 2X + 3Y= 13 (‫ب‬
X - Y = 1 ، 5X2
+ 2Y2
= 53 (‫ﺟـ‬
2X2
- Y2
= 34 ، 3X2
+ 2Y2
= 107 ( ‫د‬
/ 5‫س‬
: ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﺘﺒﺎﻳﻨﺎت‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﺟﺪ‬
( ‫ب‬ ( ‫أ‬
2X2
≤ 8 ( ‫د‬ ( ‫ﺟـ‬
3X2
-27 >0 ( ‫ﻫـ‬
A= [-2 , 5)
B = { X: X≥ 1 }
|4X + 3| =1
X |X| +4= 0
|X2
+4| = 29
|X-6| ≤ 12 ≤ |X+1| ≤4
-9 < |2X -3 |-12 ≤-3
y = 3− x+1
x x+2 = 3
2x+1 = x

404040
3 ‫ﻭﺍﻟﺠﺬﻭﺭ‬ ‫ﺍﻷﺳﺲ‬ : ‫ﺍﻟﺜﺎﻟﺚ‬ ‫ﺍﻟﻔﺼﻞ‬
‫صحيحة‬ ‫بأعداد‬ ‫األسس‬ [3-1]
‫البسيطة‬ ‫األسية‬ ‫المعادالت‬ ‫حل‬ [3-2]
‫عليها‬ ‫والعمليات‬ ‫الجذور‬ [3-3]
a + b ‫افقان‬‫ر‬‫المت‬ ‫العددان‬ [3-4]
‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻟﺪوال‬ [3-5]
-
‫●اﻷﺳﺲ‬
‫●اﻟﺠﺬور‬
‫اﻷﺳﻴﺔ‬ ‫●اﻟﺪاﻟﺔ‬
‫اﻓﻘﺎن‬‫ﺮ‬‫اﻟﻤﺘ‬ ‫●اﻟﻌﺪدان‬
a x
a + b-
‫اﻟﺴﻠﻮﻛﻴﺔ‬ ‫اﻻﻫﺪاف‬
�‫ان‬ ‫ﻋﻠﻰ‬ ً‫ا‬‫ر‬‫ﻗﺎد‬ ‫اﻟﻤﻮﺿﻮع‬ ‫ﻟﻬﺬا‬ ‫اﺳﺘﻪ‬‫ر‬‫د‬ ‫ﻧﻬﺎﻳﺔ‬ ‫ﻓﻲ‬ ‫ﻳﺼﺒﺢ‬ ‫ان‬ ‫ﻳﻨﺒﻐﻲ‬
‫ﺻﺤﻴﺤﺔ‬ ‫ﺑﺎﻋﺪاد‬ ‫اﻻﺳﺲ‬ ‫ﻗﻮاﻧﻴﻦ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ �
‫اﺳﺲ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺘﻮي‬ ‫ﺗﻤﺎرﻳﻦ‬ ‫ﻳﺤﻞ‬ �
‫ﺑﺴﻴﻄﺔ‬ ‫اﺳﻴﺔ‬ ‫ﻣﻌﺎدﻻت‬ ‫ﻳﺤﻞ‬ �
‫اﻟﺠﺬور‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ �
‫ﺟﺬور‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺘﻮي‬ ‫اﺳﺌﻠﺔ‬ ‫ﻳﺤﻞ‬ �
‫اﻓﻖ‬‫ﺮ‬‫اﻟﻤ‬ ‫اﻟﻌﺪد‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ �
‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬ ‫وﻳﺠﺪ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻟﺪوال‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ �
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺳﻠﻮك‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻌﺮف‬ �
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻳﺮﺳﻢ‬ �
‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻟﺪوال‬ ‫ﺑﻌﺾ‬ ‫ﻳﻤﺜﻞ‬ �
fa (x)=ax

41
Indices and Roots ‫واﻟﺠﺬور‬ ‫اﻷﺳﺲ‬ : ‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﻔﺼﻞ‬
: ‫ﻫﻲ‬ ‫ﺛﻼﺛﺔ‬ ‫ﺑﺎرﻛﺎن‬ ‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫ﻋﺮﻓﺖ‬ � ‫ﻣﻘﺪﻣﺔ‬
‫اﻟﺤﺴﺎب‬ ( 1
‫واﻟﻤﻘﺎﺑﻠﺔ‬ ‫اﻟﺠﺬر‬ ( 2
‫اﻟﻬﻨﺪﺳﺔ‬ ( 3
‫اﻟﺘﺎﺳﻊ‬ ‫اﻟﻘﺮن‬ ‫وﺑﺪاﻳﺔ‬ ‫ﻋﺸﺮ‬ ‫اﻟﺜﺎﻣﻦ‬ ‫اﻟﻘﺮن‬ ‫ﻧﻬﺎﻳﺔ‬ ‫ﻓﻲ‬ ‫إﻻ‬ ‫اﺑﻂ‬‫ﺮ‬‫واﻟﻤﺘ‬ ‫اﻟﻤﺘﻜﺎﻣﻞ‬ ‫ﺑﺸﻜﻠﻬﺎ‬ ‫ﺗﻌﺮف‬ ‫وﻟﻢ‬
‫وﻋﻼﻗﺘﻬﺎ‬ ‫ﺟﻬﺔ‬ ‫ﻣﻦ‬ ‫اﻟﺜﻼﺛﺔ‬ ‫اﻻرﻛﺎن‬ ‫ﺑﻴﻦ‬ ‫اﻟﻌﻼﻗﺎت‬ ‫ﻣﻦ‬ ‫اﻟﻜﺜﻴﺮ‬ ‫واﻟﻤﺴﻠﻤﻴﻦ‬ ‫اﻟﻌﺮب‬ ‫ﻋﻠﻤﺎء‬ ‫اﻛﺘﺸﻒ‬ . ‫ﻋﺸﺮ‬
: ‫وﻣﻨﻬﻢ‬ ‫اﺧﺮى‬ ‫ﺟﻬﺔ‬ ‫ﻣﻦ‬ ‫ﺑﺎﻟﻤﻨﻄﻖ‬
‫ﻓﻲ‬ ‫وﺗﻮﻓﻰ‬ ‫وﻟﺪ‬ ( ‫م‬ 1044 - 1122 ) = ( ‫ﻫـ‬ 435 - 515 ) ‫اﻟﺨﻴﺎﻣﻲ‬ ‫اﻫﻴﻢ‬‫ﺮ‬‫اﺑ‬ ‫ﺑﻦ‬ ‫ﻋﻤﺮ‬ ✳
« ‫اﻗﻠﻴﺪس‬ ‫ات‬‫ر‬‫ﻣﺼﺎد‬ ‫ﻣﻦ‬ ‫ﺷﻜﻞ‬ ‫ﻣﺎ‬ ‫ﺷﺮح‬ ‫ﻓﻲ‬ ‫رﺳﺎﻟﺔ‬ » ‫اﻟﻬﻨﺪﺳﺔ‬ ‫ﻓﻲ‬ ‫ﻛﺘﺒﻪ‬ ‫اﻫﻢ‬ ‫وﻣﻦ‬ ‫ان‬‫ﺮ‬‫اﻳ‬ ‫ﻓﻲ‬ ‫ﻧﻴﺴﺎﺑﻮر‬
. « ‫واﻟﻤﻘﺎﺑﻠﺔ‬ ‫اﻟﺠﺒﺮ‬ ‫ﻓﻲ‬ ‫ﻣﻘﺎﻟﺔ‬ » ‫ﻛﺘﺎﺑﻪ‬ ‫اﻟﺠﺒﺮ‬ ‫وﻓﻲ‬
‫ﻓﻲ‬ ‫وﻟﺪ‬ ( ‫م‬ 781 - 850 ) = ( ‫ﻫـ‬ 164 - 235 ) ‫اﻟﺨﻮارزﻣﻲ‬ ‫ﻣﻮﺳﻰ‬ ‫ﺑﻦ‬ ‫ﻣﺤﻤﺪ‬ ‫اﻟﻠﻪ‬ ‫ﻋﺒﺪ‬ ‫اﺑﻮ‬ ✳
‫اﻟﺠﺒﺮ‬ » ‫اﻟﺠﺒﺮ‬ ‫ﻓﻲ‬ ‫ﻛﺘﺒﻪ‬ ‫اﺷﻬﺮ‬ ‫وﻣﻦ‬ ‫ﺑﻐﺪاد‬ ‫اﻟﻰ‬ ‫اﻧﺘﻘﻞ‬ ‫ﺛﻢ‬ ( ً‫ﺎ‬‫ﺣﺎﻟﻴ‬ ‫اوزﺑﻜﺴﺘﺎن‬ ) ‫ﺧﻮارزم‬ ‫اﻗﻠﻴﻢ‬ ‫ﺟﻨﻮب‬
‫واﺣﺪ‬ ‫ﻣﺘﻐﻴﺮ‬ ‫ﻓﻲ‬ ‫واﻟﺜﺎﻧﻴﺔ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﻣﻦ‬ ‫ﻣﻌﺎدﻻت‬ ‫ﻟﺤﻞ‬ ‫ﺟﺒﺮﻳﺔ‬ ً‫ﺎ‬‫ﻃﺮﻗ‬ ‫اﻛﺘﺸﻒ‬ ‫اﻧﻪ‬ ‫ﻛﻤﺎ‬ « ‫واﻟﻤﻘﺎﺑﻠﺔ‬
. ‫ﻣﺘﻐﻴﺮﻳﻦ‬ ‫او‬
‫ﻧﻬﺎﻳﺔ‬ ‫ﻓﻲ‬ ‫اﻻ‬ ‫ﻣﺠﺮد‬ ‫ﺑﺸﻜﻞ‬ ‫ﺗﺪرس‬ ‫وﻟﻢ‬ ‫اﻟﻤﺠﺮد‬ ‫اﻟﺠﺒﺮ‬ ‫ﻟﻤﻮﺿﻮع‬ ‫اﻻﺳﺎس‬ ‫اﻟﺤﺠﺮ‬ ‫ﺗﻌﺘﺒﺮ‬ ‫اﻟﺰﻣﺮ‬ ‫ان‬
: ‫ﻣﻨﻬﻢ‬ ‫اﻟﺰﻣﺮ‬ ‫ﻧﻈﺮﻳﺔ‬ ‫ﺗﻄﻮﻳﺮ‬ ‫ﻓﻲ‬ ‫ﻛﺜﻴﺮون‬ ‫ﻋﻠﻤﺎء‬ ‫ﺳﺎﻫﻢ‬ ‫وﻗﺪ‬ . ‫ﻋﺸﺮ‬ ‫اﻟﺘﺎﺳﻊ‬ ‫اﻟﻘﺮن‬
‫ﻻﻧﺠﺎزه‬ ً‫ا‬‫ﺮ‬‫ﻧﻈ‬ ‫رﻳﺎﺿﻴﺎت‬ ‫ﻛﻌﺎﻟﻢ‬ ‫اﺷﺘﻬﺮ‬ ( ‫2081م‬ - 1829 ) ‫أﺑﻴﻞ‬ ، ( ‫م‬ 1777 - 1855 ) ‫ﻛﺎوس‬ ✳
‫اﻷﺑﻠﻴﺔ‬ ‫اﻟﺰﻣﺮة‬ ‫ﺑﺎﺳﻢ‬ ‫ﺗﺬﻛﺮ‬ ‫ال‬‫ﺰ‬‫وﻻﺗ‬ ‫اﺳﻤﻪ‬ ‫ﺧﻠﺪت‬ ‫اﻟﺘﻲ‬ ‫اﻻﺑﺪاﻟﻴﺔ‬ ‫اﻟﺰﻣﺮة‬ ‫واﻛﺘﺸﺎﻓﻪ‬ ‫اﻟﺰﻣﺮة‬ ‫ﻣﻮﺿﻮع‬ ‫ﻓﻲ‬
. Abelian Group

42
am
an
1
an
a
b
‫ﻋﺪد‬ ‫ﻗﻮة‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺮﻓﻨﺎ‬ ‫ﺣﻴﺚ‬ . ‫واﳉﺬور‬ ‫اﻷﺳﺲ‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫اﳌﺘﻮﺳﻄﺔ‬ ‫اﳌﺮﺣﺔ‬ ‫ﻓﻲ‬ ‫درﺳﻨﺎ‬ � ‫اﺳﺘﻪ‬‫ر‬‫د‬ ‫ﺳﺒﻘﺖ‬ ‫ﻟﻤﺎ‬ ‫ﺗﻤﻬﻴﺪ‬
‫اﳉﺬور‬ ‫ﺧﺼﺎﺋﺺ‬ ‫وﻋﻠﻰ‬ ، ‫ﺳﺎﻟﺐ‬ ‫ﻏﻴﺮ‬ ‫ﺣﻘﻴﻘﻲ‬ ‫ﻟﻌﺪد‬ ‫اﻟﺘﺮﺑﻴﻌﻲ‬ ‫اﳉﺬر‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺮﻓﻨﺎ‬ ‫ﻛﻤﺎ‬ ، ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ ‫اﻷس‬ ‫ﻳﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬
. ‫اﻟﺘﻜﻌﻴﺒﻴﺔ‬ ‫واﳉﺬور‬ ‫اﻟﺘﺮﺑﻴﻌﻴﺔ‬
‫ﺻﺤﻴﺤﺔ‬ ‫أﻋﺪاد‬ ‫اﻷﺳﺲ‬ [ 3 � 1 �
Indices ‫اﻷﺳﺲ‬
( 3 - 1 ) ‫ﺗﻌﺮﻳﻒ‬
‫ﻓﺎن‬ a∈ R ،n ∈Z ‫ﻛﺎن‬ ‫إذا‬
a0
= 1 ‫اﻟﺨﺎﺻﺔ‬ ‫اﻟﺤﺎﻟﺔ‬ ( 2
� ‫اﻷﺳﺲ‬ ‫ﺧﺼﺎﺋﺺ‬
‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ Z )
: ‫ﻓﺎن‬ b ≠ 0 ، a ≠ 0 ( ‫اﻟﺼﺤﻴﺤﺔ‬
‫ﺑﺸﺮط‬ ‫اﻻﺳﺲ‬ ‫ﺠﻤﻊ‬ُ‫ﺗ‬ ‫اﻟﻀﺮب‬ ‫ﻋﻨﺪ‬ ]an
×am
= am+n
( 1
[‫اﻷﺳﺎﺳﺎت‬ ‫ﺗﺸﺎﺑﻪ‬
a-n
= ‫ــــــ‬ ( 2
‫ﺑﺸﺮط‬ ‫اﻻﺳﺲ‬ ‫ﺗﻄﺮح‬ ‫اﻟﻘﺴﻤﺔ‬ ‫]ﻋﻨﺪ‬ ‫ــــــ‬= am-n
(3
[‫اﻟﺮﻓﻊ‬‫]ﻗﺎﻧﻮن‬ ( 4
( 5
( 6
‫اﻟﻨﻮﻧﻴﺔ‬ ‫اﻟﻘﻮة‬ an
‫اﻟﺮﻣﺰ‬ ‫ﻧﺴﻤﻲ‬
a ‫اﻟﻌﺪد‬ ‫وﻧﺴﻤﻲ‬ ، a ‫ﻟﻠﻌﺪد‬
‫وﻧﻘﻮل‬ ، ً‫ﺎ‬ّ‫ﺳ‬ُ‫أ‬ n ‫واﻟﻌﺪد‬ ً‫ﺎ‬‫أﺳﺎﺳ‬
. n ‫اﻻس‬ ‫إﻟﻰ‬ ‫ﻣﺮﻓﻮع‬ a ‫إن‬
( 1an
= a × a × . . . . . × a( ‫ﺑﻨﻔﺴﻬﺎ‬ ‫ﻣﻀﺮوﺑﺔ‬ a ‫ﻣﺮة‬ n )
1
a
a -n
= (a-1
)n
, a-1
= ‫ــــــ‬ ، a ≠ 0 ( 3
∀a ، b ∈ R ،∀n ، m∈ Z
(am
)n
= a mn
(a . b)n
= an
. bn
an
bn
‫ﻋﺪد‬ ‫ﻗﻮة‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺮﻓﻨﺎ‬ ‫ﺣﻴﺚ‬ . ‫واﳉﺬور‬ ‫اﻷﺳﺲ‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫اﳌﺘﻮﺳﻄﺔ‬ ‫اﳌﺮﺣﺔ‬ ‫ﻓﻲ‬ ‫درﺳﻨﺎ‬ � ‫اﺳﺘﻪ‬‫ر‬‫د‬ ‫ﺳﺒﻘﺖ‬ ‫ﻟﻤﺎ‬ ‫ﺗﻤﻬﻴﺪ‬
‫اﳉﺬور‬ ‫ﺧﺼﺎﺋﺺ‬ ‫وﻋﻠﻰ‬ ، ‫ﺳﺎﻟﺐ‬ ‫ﻏﻴﺮ‬ ‫ﺣﻘﻴﻘﻲ‬ ‫ﻟﻌﺪد‬ ‫اﻟﺘﺮﺑﻴﻌﻲ‬ ‫اﳉﺬر‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺮﻓﻨﺎ‬ ‫ﻛﻤﺎ‬ ، ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ ‫اﻷس‬ ‫ﻳﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬
. ‫اﻟﺘﻜﻌﻴﺒﻴﺔ‬ ‫واﳉﺬور‬ ‫اﻟﺘﺮﺑﻴﻌﻴﺔ‬
‫ﺻﺤﻴﺤﺔ‬ ‫أﻋﺪاد‬ ‫اﻷﺳﺲ‬ [ 3 � 1 �
Indices ‫اﻷﺳﺲ‬
( 3 - 1 ) ‫ﺗﻌﺮﻳﻒ‬
‫ﻓﺎن‬ a∈ R ،n ∈Z ‫ﻛﺎن‬ ‫إذا‬
a0
= 1 ‫اﻟﺨﺎﺻﺔ‬ ‫اﻟﺤﺎﻟﺔ‬ ( 2
� ‫اﻷﺳﺲ‬ ‫ﺧﺼﺎﺋﺺ‬
‫اﻷﻋﺪاد‬ ‫ﻣﺠﻤﻮﻋﺔ‬ Z )
: ‫ﻓﺎن‬ b ≠ 0 ، a ≠ 0 ( ‫اﻟﺼﺤﻴﺤﺔ‬
‫ﺑﺸﺮط‬ ‫اﻻﺳﺲ‬ ‫ﺠﻤﻊ‬ُ‫ﺗ‬ ‫اﻟﻀﺮب‬ ‫ﻋﻨﺪ‬ ]an
×am
= am+n
( 1
a-n
= ‫ــــــ‬ ( 2
‫ﺑﺸﺮط‬ ‫اﻻﺳﺲ‬ ‫ﺗﻄﺮح‬ ‫اﻟﻘﺴﻤﺔ‬ ‫]ﻋﻨﺪ‬ ‫ــــــ‬= am-n
(3
[‫اﻟﺮﻓﻊ‬‫]ﻗﺎﻧﻮن‬ ( 4
( 5
( ‫ــــــــ‬ )n
= ‫ــــــــ‬ ( 6

43
∀n ∈ N ، n > 1، n
0 = 0
Xn
Xn
=
Xn
=
Roots ‫اﻟﺠﺬور‬
( 3 - 2 ) ‫ﺗﻌﺮﻳﻒ‬
:‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻳﺤﻘﻖ‬ X ‫ﺣﻘﻴﻘﻲ‬ ‫ﻋﺪد‬ ‫ﻛﻞ‬ ‫ﻓﺄن‬ a∈R ، n∈N ، n > 1 ‫ﻛﺎن‬ ‫اذا‬
‫أو‬ n
a ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻪ‬ ‫(وﻳﺮﻣﺰ‬a)‫ﻟﻠﻌﺪد‬ً‫ﺎ‬‫ﻧﻮﻧﻴ‬ً‫ا‬‫ر‬‫ﺟﺬ‬‫ﻳﺴﻤﻰ‬Xn
=a
: ‫اﻻﺗﻴﺔ‬ ‫اﻟﻨﺘﺎﺋﺞ‬ ‫اﻟﺘﻌﺮﻳﻒ‬ ‫ﻫﺬا‬ ‫ﻣﻦ‬ ‫اﺳﺘﻨﺘﺠﻨﺎ‬ ‫ان‬ ‫ﺳﺒﻖ‬ ‫وﻗﺪ‬
( 1
‫اﻟﻌﺪدﻳﻦ‬ ‫ﻣﻦ‬ ‫ﻛﻼ‬ ‫ﻓﺎن‬ ً‫ﺎ‬‫ﻣﻮﺟﺒ‬ ً‫ﺎ‬‫ﺣﻘﻴﻘﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( a ) ‫وﻛﺎن‬ ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( n ) ‫ﻛﺎن‬ ‫اذا‬ ( 2
a ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻳﺤﻘﻖ‬ X = - n
a ،X= n
a
‫ﺣﻘﻴﻘﻲ‬ ‫ﻋﺪد‬ ‫ﻻﻳﻮﺟﺪ‬ ‫ﻓﺎﻧﻪ‬ ً‫ﺎ‬‫ﺳﺎﻟﺒ‬ ً‫ﺎ‬‫ﺣﻘﻴﻘﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( a ) ‫وﻛﺎن‬ ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( n ) ‫ﻛﺎن‬ ‫اذا‬ ( 3
(∀ X∈R‫ﻣﻮﺟﺐ‬Xn
‫ﻻن‬ ) a ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻳﺤﻘﻖ‬
‫واﺣﺪ‬ ‫ﺣﻘﻴﻘﻲ‬ ‫ﻋﺪد‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎﻧﻪ‬ ً‫ﺎ‬‫ﺣﻘﻴﻘﻴ‬ ‫ﻋﺪدا‬ ( a ) ‫وﻛﺎن‬ ً‫ﺎ‬‫ﻓﺮدﻳ‬ ً‫ﺎ‬‫ﻃﺒﻴﻌﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( n ) ‫ﻛﺎن‬ ‫اذا‬ ( 4
= a ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻳﺤﻘﻖ‬
( 3 - 1 ) ‫مبرهنة‬
‫ﻓﺎن‬ a ، b ∈ R ، n ∈ N ، n > 1 ‫ﻛﺎن‬ ‫اذا‬
( ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ا‬‫ﻋﺪد‬ ( n )‫ﻛﺎن‬ ‫اذا‬ b ≥ 0 ، a ≥ 0 ‫ﺣﻴﺚ‬ ) ، n
a.b= n
a . n
b ( 1
}n
n a
b
n
a
b
=
a
1
n
b∈R /{0} ,a∈R
‫زوﺟﻴﺎ‬ ‫ﻋﺪدا‬ n ‫ﻛﺎن‬ ‫اذا‬
‫ﻓﺮدﻳﺎ‬ ‫ﻋﺪدا‬ n ‫ﻛﺎن‬ ‫اذا‬
0 < b , 0≤a

44
� � 1 ‫ﻣﺜﺎل‬
‫ـــــــــــــــ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬
‫ــــــــــــــــــ‬ = ‫ــــــــــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــ‬
� � 2 ‫ﻣﺜﺎل‬
:‫ان‬ ‫ﻓﺎﺛﺒﺖ‬ m ، n ∈ Z ‫ﻛﺎن‬ ‫اذا‬
‫ــــــــــــــــــــــــــــــ‬ = ‫ــــــ‬
‫ـــــــــــــــــــــــــــــــ‬ = ‫ــــــــــــــــــــــــــــــــــــــــــــ‬ = ‫اﻻﻳﺴﺮ‬ ‫اﻟﻄﺮف‬
‫ــــــــــــــــــــــــــــــــــــــــ‬ =
=
‫اﻻﻳﻤﻦ‬ ‫اﻟﻄﺮف‬
8 -3
× 18 2
81 × 16 -2
8 -3
× 18 2
81 × 16 -2
( 2 3
)- 3
× ( 32
× 2 )2
34
×( 24
)-2
2 -9
× 2 2
× 3 4
3 4
× 2 -8
5
9
5 3
× ( 5 × 3 )m-2
× ( 5 2
)m+n
( 3 × 5 2
)m
× 5 2n+m
5 3
× 5 m-2
× 3 m-2
× 5 2m+2n
3 m
× 5 2m
× 5 2n+m
5
9
1
3 2
m‫ﻛﺎن‬‫اذا‬...
.ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ا‬‫ﻋﺪد‬
.ً‫ﺎ‬‫ﻓﺮدﻳ‬ ً‫ا‬‫ﻋﺪد‬ m ‫ﻛﺎن‬ ‫اذا‬ ... - a
m
‫ﻣﺜﻞ‬
(- a)m
= a
m
(-1)25
=-1
(-1)64
= 1
125 × 15m-2
× 25m+n
75m
× 52n+ m
5 3+m-2+2m+2n-2m-2n-m
× 3m-2-m
= 5 × 3-2
= 5 × ‫ــــــ‬ = ‫ــــــ‬ =
3 4-4
× 2-9+2+8
= 3 0
× 2 1
= 1 × 2 =2
×15m-2
× 25m+n
75m
× 5 2n+ m
125

45
/ 1‫س‬
:‫ﻣﺎﻳﻠﻲ‬ ‫ﻧﺎﺗﺞ‬ ‫ﺟﺪ‬
3
64 ( ‫د‬ 16 + (16)- 1
( ‫ﺟـ‬ (3) +(2)-1
( ‫ب‬ (9)0
+(8)0
( ‫أ‬
،3a0
(‫ح‬ ( 27 ) (‫ز‬ ‫ـــــــــــــــــ‬ ( ‫و‬ ‫ـــــــــــــــــ‬ ( ‫ﻫـ‬
( )-3
( ‫ك‬ (‫ي‬ ( ‫ط‬
/ 2‫س‬
: ‫ﺻﻮرة‬ ‫ﺑﺎﺑﺴﻂ‬ ‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻘﺎدﻳﺮ‬ ‫اﻛﺘﺐ‬
[‫ــــــــــــــــــــــــــــ‬ ] 2
( ‫ب‬ (‫ـــــــ‬ )2
‫ــــــــ‬ ( ‫أ‬
، ( ‫ﺟـ‬‫ــــــــــــــ‬ ، x ≠ 0 ( ‫د‬ c ≠ 0
/ 3‫س‬
ً‫ﺎ‬‫ﻣﺴﺘﺨﺪﻣ‬ ‫اﻟﺠﺬر‬ ‫ﺗﺤﺖ‬ ‫وﻻﻳﻜﻮن‬ ( 1 ) ‫ﻓﻴﻬﺎ‬ ‫اﻟﻤﻘﺎم‬ ‫ﻳﻜﻮن‬ ‫ﺑﺸﻜﻞ‬ ‫اﻻﺗﻴﺔ‬ ‫ات‬‫ر‬‫اﻟﻌﺒﺎ‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫اﻛﺘﺐ‬
: ‫اﻻﺳﺲ‬
5
( ‫ﺟـ‬ ‫ـــــــ‬ ، b ≠ 0 ( ‫ب‬ ‫ـــــــــ‬ ، d ≠ 0 ( ‫أ‬
x × 4
x ، x ≥ 0 ( ‫و‬ ‫ــــــــــــــ‬ ( ‫ﻫـ‬ ‫ـــــــــــ‬ ، b ≠ 0 ( ‫د‬
2 -3
× 4 -5
6 -1
× 3 3
10 3
× 4 7
10 -5
× 2 5
3
4
20 a 3
45
( -a )3 6
729
3 a
3x-5
. y2
2-1
y-2
b c
d
1
b 5
4b2
b2
c-3
1
b2
+ c2
-1
5‫ـــــ‬
3
( 3 a )0
،( a + b )0
a
( - a )4
25 b2
c-8
x
3
5 -32
a ≠ 0( )
a +b≠ 0( )
a ≠ 0( )
a +b ≠ 0( )

46
/ 4‫س‬
‫ﺻﺎﺋﺒﺔ؟‬ ‫ﻳﺎﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻓﺎي‬ ً‫ﺎ‬‫زوﺟﻴ‬ ً‫ﺎ‬‫ﺻﺤﻴﺤ‬ ً‫ا‬‫ﻋﺪد‬ m ‫وان‬ a∈ R ‫ﻛﺎن‬ ‫اذا‬
a m
≤ 0 (‫د‬ a m
≥ 0 (‫ﺟـ‬ a m
< 0 (‫ب‬ a m
> 0 ( ‫أ‬
/ 5‫س‬
‫ﺻﺎﺋﺒﺔ؟‬ ‫ﻳﺎﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻓﺎي‬ ً‫ﺎ‬‫ﻓﺮدﻳ‬ ً‫ﺎ‬‫ﺻﺤﻴﺤ‬ ً‫ا‬‫ﻋﺪد‬ ‫,وان‬ ‫ﺳﺎﻟﺐ‬ ‫ﻋﺪد‬ a ، a∈ R ‫ﻛﺎن‬ ‫اذا‬
a m
≤ 0 (‫د‬ a m
≥ 0 (‫ﺟـ‬ a m
< 0 (‫ب‬ a m
> 0 ( ‫أ‬
/ 6‫س‬
a (x-y)z
. a (z-x)y
. a (y-z)x
= 1 ( ‫:أ‬ ‫ان‬ ‫ﺑﺮﻫﻦ‬
(‫ب‬
/ 7‫س‬
‫ـــــــــــــــ‬ + ‫ـــــــــــــــ‬ =1 : ‫ان‬ ‫ﺑﺮﻫﻦ‬
/ 8‫س‬
‫ــــــــــــــــــــــــــــــ‬ = ‫ــــــــ‬ : ‫ان‬ ‫اﺛﺒﺖ‬
/ 9‫س‬
: ‫ﺻﻮرة‬ ‫اﺑﺴﻂ‬ ‫اﻟﻰ‬ ‫ﻳﺎﺗﻲ‬ ‫ﻣﻤﺎ‬ ً‫ﻼ‬‫ﻛ‬ ‫اﺧﺘﺼﺮ‬
‫ـــــــــــــــــــــــــــــ‬ ، ‫ــــــــــــــــــ‬
/ 10‫س‬
[ ‫ــــــــــــــــــــــــــــــ‬ ] =27 : ‫ان‬ ‫ﺑﺮﻫﻦ‬
1
1 + a b - c
1
1 + a c - b
5 × 3 2n
- 4 × 3 2 n - 1
2 × 3 2 n + 1
-32 n
11
15
3 2 +n
+ 3n + 1
3 n
-3 n - 1
6 4 n - 1
× 272 n
2 n + 1
× 8 n - 1
× 9 n + 2
( 9n +
) × 3 × 3 n
3 3 -n
1
4
m
1
n
xn2
−1
÷ xn−1⎡
⎣
⎤
⎦
1
n
= xn−1

47
‫اﻟﺒﺴﻴﻄﺔ‬ ‫اﻻﺳﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ [ 3 � 2 �
‫اﻟﻤﻌﺎدﻻت‬ ‫ﻣﻦ‬ ‫اﻟﻨﻮع‬ ‫ﻫﺬا‬ ‫وﻟﺤﻞ‬ .‫اﻻس‬ ‫ﻓﻲ‬ ‫ﻣﺘﻐﻴﺮ‬ Exponential Equation ‫اﻻﺳﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺗﺘﻀﻤﻦ‬
:‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻼﺣﻈﺎت‬ ‫ﻧﺪرج‬
((1 ≠ ‫اﻻﺳﺎس‬ ‫ﺑﺸﺮط‬ ‫اﻻﺳﺲ‬ ‫ﺗﺘﺴـــــﺎوي‬ ‫ﻓﺴﻮف‬ ‫اﻻﺳﺎﺳﺎت‬ ‫ﺗﺴﺎوت‬ ‫اذا‬ )) :‫ﻣﻌﺎدﻟﺔ‬ ‫اي‬ ‫ﻓﻲ‬ (1
‫ﻛﺎن‬ ‫اذا‬ : ‫اي‬
‫ﻓﺮدﻳﺔ‬ n ‫ﻛﺎﻧﺖ‬ ‫اذا‬ x = y ‫ﻓﺎن‬ xn
= yn
‫ﻛﺎن‬ ‫اذا‬ (2
‫زوﺟﻴﺔ‬ n ‫ﻛﺎﻧﺖ‬ ‫اذا‬ x = + y
‫ﺻﻔﺮ‬ = m = n ⇐ ‫ﻛﺎن‬ ‫اذا‬ (3
: ‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﻣﻦ‬ ‫ﻛﻼ‬ ‫ﺣﻞ‬ ‫ﻻﺣﻆ‬ ‫اﻻن‬ ‫و‬
( ‫أ‬
x + 2 = 3 ⇒ x = 1 ⇒
{1} = ‫∴ﻣﺞ‬
(‫ب‬
-
a x
= a y
⇒ x = y , a ≠ 1
xn
= ym
1
5
27
3
5
3
5
3
5( x + 2)-
= ‫ـــــــــ‬ ⇒( x + 2)-
= 3 -
x
2
3
= 3-2
x
2
3
=
1
32
(x
1
3
)2
= (
1
3
)2
‫اﻟﻄﺮﻓﻴﻦ‬ ‫ﺑﺠﺬر‬
‫اﻟﻄﺮﻓﻴﻦ‬ ‫ﺑﺘﻜﻌﻴﺐ‬x
1
3
= ±
1
3
x
1
3
)3
= ± (
1
3
)3
x= ±
1
33
x= ±
1
27

x
2
3
= 3−2
(x
2
3
) = m(3−2
)
3
2
x= m 3-3
x=
1
33
=
1
27
or x=
-1
33
=
−1
27
{± } = ‫∴ﻣﺞ‬
(

48
� � 3 ‫ﻣﺜﺎل‬
2x -2x+1
= 4x+3
: ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
‫اﻟﻤﻌﺎدﻟﺔ‬‫ﻃﺮﻓﻲ‬‫ﻓﻲ‬‫ﻧﻔﺴﻪ‬‫اﻻﺳﺎس‬‫ﻧﺠﻌﻞ‬
x2
- 2x +1 = 2x + 6 ∴
x2
-4x - 5 = 0
{-1 , 5 } = ‫اﻟﺤﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ∴
.‫ﻣﺘﻐﻴﺮة‬ ‫اﻻس‬ ‫ﻻن‬ ‫اﻻﺳﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫﺬه‬ ‫ﻣﺜﻞ‬ ‫وﺗﺴﻤﻰ‬
� � 4 ‫ﻣﺜﺎل‬
32x+1
- 4×3x+2
= -81 :‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
32x
× 3 - 4 × 3x
× 32
+81 = 0 ÷ 3
(3x
- 3) ( 3x
- 9) = 0
3x
= 9 ⇒ 3x
= 3 2
⇒ x = 2 ‫اﻣﺎ‬
3x
= 3 ⇒ x = 1 ‫او‬
. { 1 , 2 } = ‫اﻟﺤﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬
2x -2x+1
= 2 2(x+3)
( x-5) ( x+1) = 0 ⇒ x = 5 ، x = -1
32x
- 12 × 3x
+ 27 = 0
2
2

49
� � 5 ‫ﻣﺜﺎل‬
- : ‫ﻛﺎن‬ ‫اذا‬ x ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬
(‫ﺟـ‬ (‫ب‬ 3x-1
= 5x-1
( ‫أ‬
3 ‫ﻣﻼﺣﻈﺔ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ ( ‫أ‬
3x-1
= 5x-1
⇒ x - 1 = 0 ⇒ x = 1
2 ‫ﻣﻼﺣﻈﺔ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ (‫ب‬
(x+3)5
= 45
⇒ x+ 3 = 4 ⇒ x = 1
2 ‫ﻣﻼﺣﻈﺔ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ (‫ﺟـ‬
(x-1)6
= 26
⇒ x - 1 = + 2 ⇒ x = 3
x =-1
� � 6 ‫ﻣﺜﺎل‬
‫ﺣﻴﺚ‬ R ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
-
(x+3)5
= 5
4( x - 1 )6
= 2 6
8
x
2
+8
x
2
+
1
3
+8
x
2
+8
x
2
+
2
3
=14
8
x
2
+8
x
2
×8
1
3
+8
x
2
×8
2
3
=14
8
x
2
1+ 8
1
3
+8
2
3
⎛
⎝
⎜
⎞
⎠
⎟ =14
8
x
2
1+ 2+ 4( ) =14
8
x
2
× 7 =14 ⇒ 8
x
2
= 2 ⇒ 23
( )
x
2 = 2 ⇒ 8
3
x
2
= 21
⇒
3x
2
= 1⇒ x =
2
3
8
x
2
+8
x
2
+
1
3
+8
x
2
+8
x
2
+
2
3
=14
8
x
2
+8
x
2
×8
1
3
+8
x
2
×8
2
3
=14
8
x
2
1+8
1
3
+8
2
3
⎛
⎝
⎜
⎞
⎠
⎟ =14
8
x
2
1+2+ 4( ) =14
8
x
2
× 7 =14 ⇒ 8
x
2
= 2 ⇒ 23
( )
x
2 = 2 ⇒ 8
3
x
2
= 21
⇒
3x
2
= 1⇒ x =
2
3
8
x
2
+8
x
2
+
1
3
+8
x
2
+8
x
2
+
2
3
=14
8
x
2
+8
x
2
×8
1
3
+8
x
2
×8
2
3
=14
8
x
2
1+ 8
1
3
+8
2
3
⎛
⎝
⎜
⎞
⎠
⎟ =14
8
x
2
1+2+ 4( ) =14
8
x
2
× 7 =14 ⇒ 8
x
2
= 2 ⇒ 23
( )
x
2 = 2 ⇒ 8
3
x
2
= 21
⇒
3x
2
= 1⇒ x =
2
3
8
x
2
+8
x
2
+
1
3
+8
x
2
+8
x
2
+
2
3
=14
8
x
2
+8
x
2
×8
1
3
+8
x
2
×8
2
3
=14
8
x
2
1+ 8
1
3
+8
2
3
⎛
⎝
⎜
⎞
⎠
⎟ =14
8
x
2
1+2+ 4( ) =14
8
x
2
× 7 =14 ⇒ 8
x
2
= 2 ⇒ 23
( )
x
2 = 2 ⇒ 8
3
x
2
= 21
⇒
3x
2
= 1⇒ x =
2
3
8
x
2
+8
x
2
+
1
3
+8
x
2
+8
x
2
+
2
3
=14
8
x
2
+8
x
2
×8
1
3
+8
x
2
×8
2
3
=14
8
x
2
1+8
1
3
+8
2
3
⎛
⎝
⎜
⎞
⎠
⎟ =14
8
x
2
1+2+ 4( ) =14
8
x
2
× 7 =14 ⇒ 8
x
2
= 2 ⇒ 23
( )
x
2 = 2 ⇒ 8
3
x
2
= 21
⇒
3x
2
= 1⇒ x =
2
3

50
( 3 - 2 ) ‫تمرينات‬
/ 1‫س‬
:‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫ﺣﻞ‬
(‫ﺟـ‬ (‫ب‬ x3
= ‫ــــــ‬ ( ‫أ‬
(‫و‬ (‫ﻫـ‬ (‫د‬
(‫ط‬ (‫ح‬ xx-5x+6
= 1 (‫ز‬
/ 2‫س‬
‫ﺣﻴﺚ‬ R ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
/ 3‫س‬
:‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
‫ـــــــــــــــــــــــــــ‬ = 81
/ 4‫س‬
: ‫ﻋﻠﻤﺖ‬ ‫اذا‬ x∈ R ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬
3x-1
+3x
+3x +1
= 39 ( ‫أ‬
( ‫ب‬
1
2
3
x
1
27
1
2
1
2
(243)x-1
×(27)x-2
(729) x
1
2
5
( 5
243 )2
= (x-
)2
(x+2) = 3
10(x-4)(x-5)
= 1006x -3x-2
= 36 -6×5x
+25x
+5 = 0
22x+3
-57 = 65(2x
-1)5 (5x
+5-x
)=26
3 x+1
× 9 x
- 9 × 3 = 0
2
2
2 2 2
4x
+ 4 2x
( )+ 3
4x
+ 2x
= 25

51
21
3
35
n n n
55
21
3
3x
2y
3
‫ـــــــــ‬ = ‫ـــــــــ‬
3
3 x
3
2 y
5 ، 3
12 ، 6
147
4 4 4
x
y
x
y
n
n
n
‫ﻋﻠﻴﻬﺎ‬ ‫واﻟﻌﻤﻠﻴﺎت‬ ‫اﻟﺠﺬور‬ [ 3 � 3 �
:‫ﻣﺜﻞ‬ ‫ﻣﻀﺒﻮﻃﺔ‬ ‫ﺑﺼﻮرة‬ ‫ﻗﻴﻤﻬﺎ‬ ‫اﻳﺠﺎد‬ ‫ﻳﻤﻜﻦ‬ ‫ﻻ‬ ‫ﻛﻤﻴﺎت‬ ‫ﻫﻲ‬ ‫اﻟﺠﺬور‬ ‫ﺑﻌﺾ‬
61 ، 10 ، 2
. ‫ﺗﺒﺴﻴﻄﻬﺎ‬ ‫ﻋﻤﻠﻴﺔ‬ ‫ﻟﺘﺴﻬﻴﻞ‬ ‫اﻟﺨﻮاص‬ ‫ﺑﻌﺾ‬ ‫وﺳﻨﻌﻄﻲ‬ ‫اﻟﺼﻤﺎء‬ ‫ﺑﺎﻟﺠﺬور‬ ‫اﻟﺠﺬور‬ ‫ﻫﺬه‬ ‫ﺗﺪﻋﻰ‬
‫اﻟـﺨــــــﻮاص‬
.‫ﺻﺤﻴﺢ‬‫اﻟﺨﺎﺻﻴﺔ‬‫وﻋﻜﺲ‬ x × y = xy .1
6 × 12 = 72 : ً‫ﻼ‬‫ﻣﺜ‬
5 3 x3
= 4
15x3
.y ≠ o ‫ﺣﻴﺚ‬ ‫ﺻﺤﻴﺢ‬ ‫اﻟﺨﺎﺻﻴﺔ‬ ‫وﻋﻜﺲ‬ = ‫ـــــــــ‬
‫ــــــــ‬ = ‫ــــــــ‬ = 7 : ً‫ﻼ‬‫ﻣﺜ‬
� � 7 ‫ﻣﺜﺎل‬
:ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﻳ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﺠﺬور‬ ‫رﺗﺐ‬
6
147 ، 5 ، 3
12
3
12 = 6
12 2
= 6
144
5 = 6
53
= 6
125
6
147 = 6
147
: ‫ﻳﻜﻮن‬ ‫اﻟﺘﺮﺗﻴﺐ‬
5

52
Conjugate Numbers ‫اﻓﻘﺎن‬‫ﺮ‬‫اﻟﻤﺘ‬ ‫اﻟﻌﺪدان‬ [ 3 � 4 �
. ‫ﻧﺴﺒﻴﺔ‬ ‫ﻛﻤﻴﺔ‬ ‫اﻟﻰ‬ ‫ﻟﺘﺤﻮﻟﺖ‬ ‫اﻟﻨﺴﺒﻴﺔ‬ ‫ﻏﻴﺮ‬ ‫اﻟﻜﻤﻴﺔ‬ ‫ﺑﻪ‬ ‫ﺿﺮﺑﺖ‬ ‫ﻟﻮ‬ ‫اﻟﺬي‬ ‫ﻫﻮ‬ ‫اﻟﻤﻨﺴﺐ‬ ‫اﻟﻌﺎﻣﻞ‬ ‫ان‬ ‫ﻧﻌﻠﻢ‬
‫ﻷن‬ 3 ‫ﻫﻮ‬ 2 3 ‫ﻟﻠﻤﻘﺪار‬ ‫اﻟﻤﻨﺴﺐ‬ ‫ﻓﺎﻟﻌﺎﻣﻞ‬
3
3 × 3
32
= 3
33
= 3 ‫ﻷن‬ 3
32 ‫ﻫﻮ‬ 3
3 ‫ﻟﻠﻤﻘﺪار‬ ‫اﻟﻤﻨﺴﺐ‬ ‫واﻟﻌﺎﻣﻞ‬
‫اﻓﻘﻪ‬‫ﺮ‬‫ﻣ‬ ‫ﻫﻮ‬ ‫ﻟﻠﻤﻘﺪار‬ ‫اﻟﻤﻨﺴﺐ‬ ‫واﻟﻌﺎﻣﻞ‬
( ) ( ) = 25 - 6 = 19 ‫ﺿﺮﺑﻬﻤﺎ‬ ‫ﻻن‬
‫ﻟﻠﻤﻘﺪار‬ ‫اﻟﻤﻨﺴﺐ‬ ‫واﻟﻌﺎﻣﻞ‬
( 3 2 - 2 5 ) ( 3 2 + 2 5 ) = 9 × 2 - 4 × 5 = -2
‫ﻫﻮ‬ 53
−1 ‫ﻟﻠﻤﻘﺪار‬ ‫اﻟﻤﻨﺴﺐ‬ ‫واﻟﻌﺎﻣﻞ‬
( 3
5 - 1) ( 3
25 + 3
5 + 1) = 3
125 - 1 = 5 - 1 = 4
( ‫ﻣﻜﻌﺒﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫ﻓﺮق‬ )
� � 8 ‫ﻣﺜﺎل‬
:‫ﻧﺴﺒﻴﺔ‬ ‫ﻛﻤﻴﺔ‬ ‫اﻟﻤﻘﺎم‬ ‫ﻳﻜﻮن‬ ‫ﺑﺤﻴﺚ‬ ‫ﺑﺴﻂ‬
‫ــــــــــــ‬ + ‫ــــــــــــــــــ‬ + ‫ـــــــــــ‬
‫ــــــــــــ‬ × ‫ــــــــــــــ‬ + ‫ـــــــــــــــــ‬ × ‫ـــــــــــــــــــــ‬ + ‫ــــــــــــــ‬ × ‫ــــــــــــــ‬
= ‫ـــــــــــــ‬ + ‫ـــــــــــــــــــ‬ + ‫ــــــــــــــ‬
= 2 + 1 + 3 - 2 + 2 - 3 = 3
-
1
2 - 1
1
2 + 3
1
3 + 2
a + b
2 3 × 3 = 2 × 3 = 6
5 + 65 - 6
5 + 6
3 2 + 2 5 3 2 - 2 5‫ﻫﻮ‬
‫ﻻن‬
5 - 6
3
52
+ 3
5 +1
‫ﻻن‬
1
2 - 1
2 + 1
2 + 1
1
3 + 2
3 - 2
3 - 2
1
2 + 3
2 - 3
2 - 3
2 + 1
2 - 1
2 - 3
4 - 3
3 - 2
3 - 2

53
Real Functions ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻟﺪوال‬ [ 3 � 5 �
‫اﻟﺪاﻟﺔ‬ ‫ﻣﻔﻬﻮم‬ ‫ﺳﻨﻮﺿﺢ‬ ‫واﻻن‬ .‫ان‬‫ﺮ‬‫اﻗﺘ‬ ‫وﻗﺎﻋﺪة‬ ‫ﻣﻘﺎﺑﻞ‬ ‫ﻣﺠﺎل‬ ،‫ﻣﺠﺎل‬ :‫ﻣﻦ‬ ‫ﻳﺘﻜﻮن‬ ‫اﻧﻪ‬ ‫وﻋﺮﻓﻨﺎ‬ ‫اﻟﺘﻄﺒﻴﻖ‬ ‫ﺳﺎﺑﻘﺎ‬ ‫درﺳﻨﺎ‬
:‫ﺑﺸﻜﻞ‬ ‫ﺗﻜﺘﺐ‬ .R ‫ﻣﻦ‬ ‫ﺧﺎﻟﻴﺔ‬ ‫ﻏﻴﺮ‬ ‫ﺟﺰﺋﻴﺔ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻘﺎﺑﻞ‬ ‫وﻣﺠﺎﻟﻬﺎ‬ ‫ﻣﺠﺎﻟﻬﺎ‬ ً‫ﺎ‬‫اﻳﻀ‬ ‫ﺗﻄﺒﻴﻖ‬ ‫ﻫﻲ‬ ‫اﻟﺘﻲ‬
y = f(x) , A,B ⊆ R ،∈ y = f(x) , A,B ⊆ R‫ان‬ ‫ﺣﻴﺚ‬ (‫وﺣﻴﺪ‬ ‫ﻋﻨﺼﺮ‬ y)f: A → B , ∀ x ∈ A ∃ y ∈ B ،f: A → B , ∀ x ∈ A ∃ y ∈ B‫ﻳﻌﻨﻲ‬ f: A → B , ∀ x ∈ A
‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫ﻟﻠﺪوال‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬ ‫اﻳﺠﺎد‬ [ 3 � 5 � 1 �
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ،‫اﻟﺠﺬرﻳﺔ‬ ‫اﻟﺪاﻟﺔ‬ ،‫اﻟﻜﺴﺮﻳﺔ‬ ‫اﻟﺪاﻟﺔ‬ ،‫اﻟﺤﺪود‬ ‫ﻛﺜﻴﺮة‬ ‫اﻟﺪاﻟﺔ‬ :‫ﻫﻲ‬ ‫اﻟﺪوال‬ ‫ﻣﻦ‬ ‫اﻧﻮاع‬ ‫ارﺑﻌﺔ‬ ‫ﻫﻨﺎ‬ ‫ﺳﻨﺪرس‬
‫اﺧﺮى‬ ‫اﻟﻰ‬ ‫داﻟﺔ‬ ‫ﻣﻦ‬ ‫ﻳﺨﺘﻠﻒ‬ ‫اﻟﻤﺠﺎل‬ ‫ﺣﻴﺚ‬
:‫ﺑﺎﻟﺸﻜﻞ‬ ‫وﺗﻜﺘﺐ‬ ‫اﻟﺤﺪود‬ ‫ﻛﺜﻴﺮة‬ ‫اﻟﺪاﻟﺔ‬ *
‫واﻟﺪاﻟﺔ‬ ،x3
+ 2x2
+x-1 , g(x) = x2
− 5x + 9 , f(x) = 3x - 1:‫ﻣﺜﻞ‬ ‫اﻟﺘﺮﺑﻴﻌﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ،f(x) = x3
+ 2x2
+x-1 , g(x) = x2
− 5x + 9 , f(x) = 3x - 1 :‫ﻣﺜﻞ‬ ‫اﻟﺨﻄﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫وﺗﺸﻤﻞ‬
f(x) = x3
+ 2x2
+x-1 , g(x) = x2
− 5x + 93) h(x) =
x + 7
x2
- 3x
⇒ x2
- 3x = 0 ⇒ x(x-3:‫ﻣﺜﻞ‬ ‫اﻟﺘﻜﻌﻴﺒﻴﺔ‬
R ‫ﻳﺴﺎوي‬ ‫اﻟﺤﺪود‬ ‫ﻛﺜﻴﺮة‬ ‫ﻟﻠﺪوال‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬
.R ‫ﻳﺴﺎوي‬ ‫ﻟﻬﺎ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬ f(x) = x3
+ 2x2
+x-1 , g(x) = x2
− 5x + 9 , f(x) = 3x - 1:ً‫ﻼ‬‫ﻣﺜ‬
x ‫ﻗﻴﻢ‬ ‫وﻧﺠﺪ‬ ً‫ا‬‫ﺮ‬‫ﺻﻔ‬ = ‫اﻟﻤﻘﺎم‬ ‫ﻧﺠﻌﻞ‬ ‫اﻟﺪوال‬ ‫ﻣﻦ‬ ‫اﻟﻨﻮع‬ ‫ﻫﺬا‬ ‫ﻣﺠﺎل‬ ‫ﻻﻳﺠﺎد‬ :‫اﻟﻜﺴﺮﻳﺔ‬ ‫اﻟﺪاﻟﺔ‬ *
:ً‫ﻼ‬‫ﻣﺜ‬ .R{x ‫}ﻗﻴﻢ‬ ‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬ ‫ﻓﻴﻜﻮن‬
1) f(x) =
2x-1
x+5
⇒ x + 5 = 0 ⇒ x = −5 ⇒ R{-5}‫ﻧﺠﻌﻞ‬1) f(x) =
2x-1
x+5
⇒ x + 5 = 0 ⇒ x = −5 ⇒ R{-5}1) f(x) =
2x-1
x+5
⇒ x + 5 = 0 ⇒ x = −5 ⇒ R{-5} ‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬

2) g(x) =
2
x2
− 4
⇒ x2
-4 = 0 ⇒ x = m 2 ⇒ R{ m 2}‫ﻧﺠﻌﻞ‬

2) g(x) =
2
x2
− 4
⇒ x2
-4 = 0 ⇒ x = m 2 ⇒ R{ m 2}

2) g(x) =
2
x2
− 4
⇒ x2
-4 = 0 ⇒ x = m 2 ⇒ R{ m 2} ‫ﻟﻠﺪاﻟﻪ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬
3) h(x) =
x + 7
x2
- 3x
⇒ x2
- 3x = 0 ⇒ x(x-3) = 0 ⇒ x = 0, x = 3 ⇒ R{0,3}‫ﻧﺠﻌﻞ‬3) h(x) =
x + 7
x2
- 3x
⇒ x2
- 3x = 0 ⇒ x(x-3) = 0 ⇒ x = 0, x = 3 ⇒ R{0,3}
x) =
x + 7
x2
- 3x
⇒ x2
- 3x = 0 ⇒ x(x-3) = 0 ⇒ x = 0, x = 3 ⇒ R{0,3} ‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬
‫داﺧﻞ‬ ‫ﺗﺠﻌﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ x ‫ﻗﻴﻢ‬ ‫ﺟﻤﻴﻊ‬ ‫ﻧﺠﺪ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫ﻻﻳﺠﺎد‬ :(‫زوﺟﻲ‬ ‫اﻟﺠﺬر‬ ‫)دﻟﻴﻞ‬ ‫اﻟﺠﺬرﻳﺔ‬ ‫اﻟﺪاﻟﺔ‬ *
:ً‫ﻼ‬‫ﻣﺜ‬ ً‫ا‬‫ﺮ‬‫ﺻﻔ‬ ‫ﻳﺴﺎوي‬ ‫او‬ ‫اﻛﺒﺮ‬ ‫اﻟﺠﺬر‬
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}x
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}-
1
2
1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
1) f(x) = x-7 , x-7 ≥ 0 ⇒ x ≥ 7 ⇒ {x ∈R:x ≥ 7}
2) g(x) = 3x+5, 3x + 5 ≥ 0 ⇒ x ≥
-5
3
⇒ {x ∈R: x ≥
-5
3
}
3) h(x) = 1-2x , 1-2 ≥ 0 ⇒ 2x ≤ -1⇒ x ≤
-1
2
⇒ {x ∈R:x ≤
-1
2
}
‫اﻻس‬ ‫ﻳﻤﺜﻞ‬ x ، ‫اﻻﺳﺎس‬ ‫ﻫﻮ‬ a ‫ﺣﻴﺚ‬ x ∈ R ، a ∈R+
{1} ‫ﺣﻴﺚ‬ fa (x)=ax
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ *
f1
2
(x) = (
1
2
)x
, h 5
(x) = ( 5)x
, g3 (x)=3x
, f2 (x) = 2x :‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻻﻣﺜﻠﺔ‬ ‫ﻣﻦ‬
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻓﻲ‬ a=1 ‫ﻧﺴﺘﺒﻌﺪ‬ ‫ﺟﻌﻠﻨﺎ‬ ‫ﻣﺎ‬ ‫وﻫﺬا‬ ‫ﺛﺎﺑﺘﺔ‬ ‫داﻟﺔ‬ ‫وﻫﺬه‬ f(x) = 1x
= 1 :‫ﻣﻼﺣﻈﺔ‬
‫ﻟﻠﺪاﻟﺔ‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬
f(x) = anxn
+ an-1xn-1
+...+ a0
an ,a1, ... a0 ∈Ran ,an-1,...a0 ∈R

54
‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫ﻟﻠﺪوال‬ ‫اﻟﺒﻴﺎﻧﻲ‬ ‫اﻟﺘﻤﺜﻴﻞ‬ [ 3 � 5 � 2 �
a ≠ 0 , a,b ∈R ‫ﺣﻴﺚ‬ f(x) = ax + b ‫اﻟﺨﻄﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺗﻤﺜﻴﻞ‬ :ً‫ﻻ‬‫او‬
f(x) = 2x + 3 , ∀ x ∈R ‫اﻟﺪاﻟﺔ‬ ‫ﻣﺜﻞ‬ :‫ﻣﺜﺎل‬
x 1 0 -1
y 5 3 1
‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﺧﻂ‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻫﺬه‬ ‫ان‬ ‫ﻻﺣﻆ‬
a,b ∈R, a ≠ 0 , f(x) = ax2
+b ‫اﻟﺘﺮﺑﻴﻌﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺗﻤﺜﻴﻞ‬ :ً‫ﺎ‬‫ﺛﺎﻧﻴ‬
a > 0, b ≥ 0 ‫ﻋﻨﺪﻣﺎ‬ ‫اي‬ f(x) = 2x2
+3 ‫اﻟﺪاﻟﺔ‬ ‫ﻣﺜﻞ‬ :‫ﻣﺜﺎل‬
x -1 0 1
y 5 3 5

U ‫ﻫﻮ‬ ‫ﻟﺪاﻟﺔ‬ ‫ﺷﻜﻞ‬ ‫ان‬ ‫ﻻﺣﻆ‬
‫اﻻﺣﺪاﺛﻲ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻣﻦ‬ ‫اﻻﻋﻠﻰ‬ ‫اﻟﻨﺼﻒ‬ ‫ﻓﻲ‬ ‫ﻳﻘﻊ‬ ‫اﻟﺒﻴﺎﻧﻲ‬ ‫وﺗﻤﺜﻴﻠﻬﺎ‬
a < 0 ‫ﻋﻨﺪﻣﺎ‬ ‫اي‬ ،f(x) = -4x2
‫اﻟﺪاﻟﺔ‬ ‫ﻣﺜﻞ‬ :‫ﻣﺜﺎل‬
x 1 0 -1
y -4 0 -4

I ‫ﻫﻮ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺷﻜﻞ‬ ‫ان‬ ‫ﻻﺣﻆ‬
‫اﻻﺣﺪاﺛﻲ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻣﻦ‬ ‫اﻻﺳﻔﻞ‬ ‫اﻟﻨﺼﻒ‬ ‫ﻓﻲ‬ ‫ﻳﻘﻊ‬ ‫اﻟﺒﻴﺎﻧﻲ‬ ‫وﺗﻤﺜﻴﻠﻬﺎ‬
‫اﻟﺘﻜﻌﻴﺒﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺗﻤﺜﻴﻞ‬ :ً‫ﺎ‬‫ﺛﺎﻟﺜ‬
a,b ∈R , a ≠ 0, f(x) = ax3
+b
f(x) = x3
+2 :‫اﻟﺪاﻟﺔ‬ ‫ﻣﺜﻞ‬ :‫ﻣﺜﺎل‬
x 1 0 -1
y 3 2 1
f(x) = -x3
‫اﻟﺪاﻟﺔ‬ ‫ﻣﺜﻞ‬ :‫ﻣﺜﺎل‬
x 1 0 -1
y -1 0 1
y
0 x
y
x
y
x
y
x
y
x
y = 2x + 3
f(x) = -x3
f(x) = x3
+2

55
fa (x)=ax
‫اﻻﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺗﻤﺜﻴﻞ‬ :ً‫ﺎ‬‫اﺑﻌ‬‫ر‬
:‫ﻣﺜﺎل‬
‫رﺳﻢ‬ ‫ﻓﻲ‬ ‫ذﻟﻚ‬ ‫ﻣﻦ‬ ‫اﺳﺘﻔﺪ‬ ‫ﺛﻢ‬ 3 ،2 ،1 ،0 ،-1 ،-2 ،-3 ‫اﺟﻞ‬ ‫ﻣﻦ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﻗﻴﻢ‬ ‫ﺟﺪ‬ ( ‫أ‬
.‫اﻟﺪاﻟﺔ‬ ‫ﻫﺬه‬ ‫ﻣﻨﺤﻨﻲ‬ ‫ﻣﻦ‬ ‫ﺟﺰء‬
(f (x :‫اﻟﺪاﻟﺔ‬ ‫ﻫﺬه‬ ‫ﻣﻨﺤﻨﻲ‬ ‫ﻣﻦ‬ ‫ﺟﺰء‬ ‫رﺳﻢ‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫ﻣﻦ‬ ‫ﻟﻼﻓﺎدة‬ ‫ﻃﺮﻳﻘﺔ‬ ‫ﻋﻦ‬ ‫اﺑﺤﺚ‬ (‫ب‬
.‫ﻧﻔﺴﻪ‬ ‫اﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬
:‫اﻟﺤﻞ‬
f (x) = 2x
( ‫أ‬
-3-2-10123x
‫ــــــ‬‫ــــــ‬12482x 1
2
1
4
1
8
‫ــــــ‬
‫اﻟﺼﺎدات‬ ‫ﻟﻤﺤﻮر‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ ‫ﺗﻨﺎﻇﺮ‬RY
‫وﻟﻨﻔﺮض‬ ( ‫ب‬
‫ﺻــــــﻮرة‬ ‫ان‬ ‫اي‬
‫ﻓــﺎﻧﻨــــــﺎ‬ ‫ﻟــﺬﻟﻚ‬ (x , 2 x
) = (-x , 2x
)
‫ﻟﺪاﻟﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬
‫ﺣـﻮل‬ ‫ﺑﺎﻟﺘﻨﺎﻇﺮ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫ﻣــﻦ‬
‫ﻓـــﻲ‬ ‫ﻣﻮﺿــﺢ‬ ‫ﻛﻤـــﺎ‬ ‫اﻟﺼــﺎدات‬ ‫ﻣﺤــﻮر‬
(3 - 1) ‫اﻟﺸﻜﻞ‬
(3 - 1) ‫اﻟﺸﻜﻞ‬
1
2
f (x) = 2xx =
f (x)
1
2
g (x) = f (x) =( ‫ــــــ‬ )x
= (2-1
)x
= 2- x
= f(-x)
Ry
: (x, y) = (-x, y)
1
2g(x)=( )x
f (x) = 2x
1
2
‫ﻓــﺎﻧﻨــــــﺎ‬ ‫ﻟــﺬﻟﻚ‬ (
‫ﺣـﻮل‬ ‫ﺑﺎﻟﺘﻨﺎﻇﺮ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫ﻣــﻦ‬
‫ﻓـــﻲ‬ ‫ﻣﻮﺿــﺢ‬ ‫ﻛﻤـــﺎ‬ ‫اﻟﺼــﺎدات‬ ‫ﻣﺤــﻮر‬
1 2 3
2 x
2 -x
y
x
x
x
x
x
x
x
x
x x x
xx
-3 -2 -1

56
1
2
( ‫ــــــ‬ )x
( ‫ــــــ‬ )x1
5
1
3
( ‫ــــــ‬ )x 1
4
( ‫ــــــ‬ )x
، ، ، ،.......
o
y
x
(0,1)
2x
3x
4x
1
2
( ‫ــــ‬ )x
1
3
( ‫ــــ‬ )x
1
4
( ‫ــــ‬ )x
:f (x) = ax
‫اﻷﺳﻴﺔ‬ ‫اﻟﺪاﻟﺔ‬ ‫ﺧﺼﺎﺋﺺ‬ ‫ﺑﻌﺾ‬
: ‫اﻟﺪوال‬ ‫ﻣﻨﺤﻨﻴﺎت‬ ‫ﺑﺮﺳﻢ‬ ‫ﻗﻤﻨﺎ‬ ‫اذا‬ .1
2x
، 3x
، 4x
، 5x
، ......
: ‫اﻟﺪوال‬ ‫وﻛﺬﻟﻚ‬
:‫اﻟﻤﻨﺤﻨﻴﺎت‬ ‫ﻣﻦ‬ ‫ﻣﺠﻤﻮﻋﺘﻴﻦ‬ ‫ﻧﺠﺪ‬ ‫ﻓﺴﻮف‬
. x ‫ﻗﻴﻤﺔ‬ ‫اﻳﺪت‬‫ﺰ‬‫ﺗ‬ ‫ﻛﻠﻤﺎ‬ ax
‫اﻟﺪاﻟﺔ‬ ‫ﻗﻴﻢ‬ ‫اﻳﺪ‬‫ﺰ‬‫ﺗﺘ‬ ‫ﺣﻴﺚ‬ a > 1 ‫ﻋﻨﺪﻣﺎ‬ : ‫اﻻوﻟﻰ‬
. x ‫ﻗﻴﻤﺔ‬ ‫اﻳﺪت‬‫ﺰ‬‫ﺗ‬ ‫ﻛﻠﻤﺎ‬ ax
‫اﻟﺪاﻟﺔ‬ ‫ﻗﻴﻢ‬ ‫ﺗﺘﻨﺎﻗﺺ‬ ‫ﺣﻴﺚ‬ 1> > 0 ‫ﻋﻨﺪﻣﺎ‬ : ‫اﻟﺜﺎﻧﻴﺔ‬
(‫ﻣﻨﺤﻨﻲ‬ ‫ﻛﻞ‬ ‫ﻣﻦ‬ ‫ﺟﺰء‬ ‫رﺳﻢ‬ ) ‫اﻟﻤﻨﺤﻨﻴﺎت‬ ‫ﻫﺬه‬ ‫ﻣﻦ‬ ‫ﺳﺘﺔ‬ ( 3 - 2 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫رﺳﻤﻨﺎ‬ ‫وﻗﺪ‬
‫ﻣﻘﻠﻮﺑﺎت‬ ‫اﻻﺧﻴﺮة‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ a ‫ﻗﻴﻢ‬ ‫اﺧﺘﺮﻧﺎ‬ ‫وﻗﺪ‬ 1 > > 0 ‫ﻓﻴﻬﺎ‬ ‫اﺧﺮى‬ ‫ﻣﻨﻬﺎ‬ ‫وﺛﻼﺛﺔ‬ a > 1 ‫ﻓﻴﻬﺎ‬ ‫ﺛﻼﺛﺔ‬
(0 , 1) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫ﺗﻤﺮ‬ ‫اﻟﻤﻨﺤﻨﻴﺎت‬ ‫ﻫﺬه‬ ‫ﺟﻤﻴﻊ‬ ‫ان‬ ‫وﻧﻼﺣﻆ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺜﻼﺛﺔ‬ ‫ﻓﻲ‬ a ‫ﻗﻴﻢ‬
. R ‫ﻣﺠﺎﻟﻬﺎ‬ ‫ان‬ ‫ﻧﺠﺪ‬ a ≠ 0 ، ax
‫اﺳﻴﺔ‬ ‫داﻟﺔ‬ ‫ﻷﻳﺔ‬ ‫اﻟﺒﻴﺎﻧﻲ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫اﻟﻰ‬ ‫ﺑﺎﻟﺮﺟﻮع‬ .2
( 3 - 2 ) ‫اﻟﺸﻜﻞ‬
a
a

57
/ 1‫س‬
/ 3‫س‬
/4‫س‬:‫ﻧﺴﺒﻴﺔ‬ ‫ﻛﻤﻴﺔ‬ ‫اﻟﻤﻘﺎم‬ ‫ﻳﻜﻮن‬ ‫ﺑﺤﻴﺚ‬ ‫ﻣﺎﻳﻠﻲ‬ ‫ﻧﺎﺗﺞ‬ ‫اوﺟﺪ‬
( ‫أ‬
(‫ب‬
( ‫أ‬
(‫ﺟـ‬
‫ﺑﺼﻮرة‬ x ‫ﺟﺪ‬ (‫ب‬
‫ان‬ ‫اﺛﺒﺖ‬ / 5‫س‬
: ‫ﻛﺎﻧﺖ‬ ‫اذا‬
. :‫ﻟﻠﺪاﻟﺔ‬ ‫اﻟﺒﻴﺎﻧﻲ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫ﻣﻦ‬ ً‫ا‬‫ﺟﺰء‬ ‫ارﺳﻢ‬ / 6‫س‬
2x
a - b
18x3
( a - b )5
‫ـــــــــ‬ ‫ــــــــــ‬ . ‫ــــــ‬ ÷ ‫ــــــــــــ‬
5 - 1
- 15
x - 6 + x
5
24
‫ـــــــ‬ 2 ( 3 + ‫ـــــــ‬ )
Ans :
3
2
- ‫ــــــ‬‫ـــــ‬ 8
27‫ــــــــــــــــــــــــــ‬
3
x - 2
‫ـــــــــــــــــ‬ + ‫ــــــــــ‬ - ‫ــــــــــ‬ = 03
x +3
‫ــــــــــــــ‬ - ‫ـــــــــــــ‬5 + 1 5 - 1
5 + 1
3
a - b
a - b
x
Ans :
1
3
Ans : 1
a + 3b
x+ 3x = 8
y =
1
5
⎛
⎝
⎜
⎞
⎠
⎟
x
a + 3b
x+ 3x = 8
y =
1
5
⎛
⎝
⎜
⎞
⎠
⎟
x
‫ـــــــــــــــ‬ ‫ــــــــــــــــ‬ - ‫ــــــــــــــــ‬
-1
‫اﺧﺘﺼـــــﺮ‬ (‫أ‬a 2
- b2
a + b ]] a - b
a + b
a + b
a - b
‫ﻛﺎن‬ ‫اذا‬ (‫ب‬
‫ان‬ ‫ﻓﺎﺛﺒﺖ‬
y = 43
− 2 +1
, x = 23
+1
x y = 3
x y =
:‫اﻟﺘﺎﻟﻴﺔ‬ ‫ﻟﻠﺪوال‬ ‫ﻣﺠﺎل‬ ‫اوﺳﻊ‬ ‫ﺟﺪ‬
a) f(x) = x2
− 5+9 b) f(x) =
x-1
x+9
c) f(x) = x-9
d) f(x) = 3-5x e) f(x) =
1
x2
− 9
:‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺪوال‬ ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ‫ﻣﺜﻞ‬ /:2‫س‬
a) f(x) = -4x2
+ 5 b) f(x) = x - 8 c) f(x) = 2 - x3

585858
4
‫ﺍﻟﻤﺜﻠﺜﺎﺕ‬ ‫ﺣﺴﺎﺏ‬ : ‫ﺍﻟﺮﺍﺑﻊ‬ ‫ﺍﻟﻔﺼﻞ‬
‫القياسي‬ ‫بالوضع‬ ‫الموجهة‬ ‫اوية‬‫ز‬‫ال‬ [4-1]
‫للزوايا‬ ‫الدائري‬ ‫والقياس‬ ‫الستيني‬ ‫القياس‬ [4-2]
‫للزوايا‬ ‫والدائري‬ ‫الستيني‬ ‫القياس‬ ‫بين‬ ‫العالقة‬ [4-3]
‫االساسية‬ ‫العالقات‬ ‫وبعﺾ‬ ‫حادة‬ ‫لزوايا‬ ‫المثلثية‬ ‫النسبة‬ [4-4]
‫ﺧﺎﺻﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺐ‬ [4-5]
‫المثلثية‬ ‫والنقطة‬ ‫الوحدة‬ ‫دائرة‬ [4-6]
‫الدائرية‬ ‫التطبيقات‬ [4-7]
‫الدائرية‬ ‫التطبيقات‬ ‫قيم‬ ‫ايجاد‬ ‫في‬ ‫الحاسوب‬ ‫استخدام‬ [4-8]
‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﺣﻞ‬ [4-9]

( B A , B C ) ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ✳
D° ، Q ‫واﻟﺪاﺋﺮي‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ✳
Sin x x ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﺟﻴﺐ‬ ✳
Cos x x ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﺗﻤﺎم‬ ‫ﺟﻴﺐ‬ ✳
Tan x x ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻇﻞ‬ ✳
( Cos x , Sin x ) ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﻘﻄﺔ‬ ✳
←←
:‫ان‬ ‫على‬ ً‫ا‬‫ر‬‫قاد‬ ‫الطالب‬ ‫يكون‬ ‫الفصل‬ ‫هذا‬ ‫اسة‬‫ر‬‫د‬ ‫بعد‬ ‫ينبﻐي‬
‫الموجهة‬ ‫اوية‬‫ز‬‫ال‬ ‫على‬ ‫يتعرف‬ -
‫الدائري‬ ‫والنظام‬ ‫الستيني‬ ‫النظام‬ ‫على‬ ‫يتعرف‬ -
‫النظامين‬ ‫بين‬ ‫يميز‬ -
‫المثلثية‬ ‫النسب‬ ‫على‬ ‫يتعرف‬ -
‫االساسية‬ ‫العالقات‬ ‫بعﺾ‬ ‫على‬ ‫يتعرف‬ -
‫الخاصة‬ ‫للزوايا‬ ‫المثلثية‬ ‫النسب‬ ‫على‬ ‫يتعرف‬ -
‫المثلثية‬ ‫النسب‬ ‫على‬ ‫تعتمد‬ ‫مسائل‬ ‫يحل‬ -
‫الوحدة‬ ‫دائرة‬ ‫على‬ ‫يتعرف‬ -
‫المثلثية‬ ‫النقطة‬ ‫على‬ ‫يتعرف‬ -
‫الرياضية‬ ‫العمليات‬ ‫بعﺾ‬ ‫في‬ ‫الحاسوب‬ ‫يستخدم‬ -
‫اوية‬‫ز‬‫ال‬ ‫القائم‬ ‫المثلث‬ ‫يحل‬ -

59
: Trigonometry ‫اﻟﻤﺜﻠﺚ‬ ‫ﺣﺴﺎب‬ : ‫اﺑﻊ‬‫ﺮ‬‫اﻟ‬ ‫اﻟﻔﺼﻞ‬
‫ﻛﺘﺐ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﻋﻠﻢ‬ ‫ﻣﻦ‬ ‫ﺗﺸﺘﺖ‬ ‫ﻣﺎ‬ ‫وﺟﻤﻊ‬ ‫ﺗﻌﺪﻳﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻜﺒﻴﺮ‬ ‫اﻟﻔﻀﻞ‬ ‫ﻟﻠﻤﺴﻠﻤﻴﻦ‬ ‫ان‬ � ‫ﻣﻘﺪﻣﺔ‬
.‫اﻟﻌﻠﻢ‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫واﺿﺤﺔ‬ ‫ات‬‫ر‬‫اﺷﺎ‬ ‫واﻟﻴﻮﻧﺎﻧﻴﻦ‬ ‫واﻟﺼﻴﻨﻴﻦ‬ ‫واﻟﻬﻨﻮد‬ ‫واﻟﻤﺼﺮﻳﻴﻦ‬ ‫ﻟﻠﺒﺎﺑﻠﻴﻴﻦ‬ ‫ﻓﻜﺎن‬ . ‫اﻻﻏﺮﻳﻖ‬
: ‫اﻟﻤﺠﺎل‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫اﺳﻬﻤﻮا‬ ‫اﻟﺬﻳﻦ‬ ‫اﻟﻤﺴﻠﻤﻴﻦ‬ ‫اﻟﻌﻠﻤﺎء‬ ‫وﻣﻦ‬
‫اﻟﻔﻠﻜﻲ‬ ‫اﺣﻤﺪ‬ ‫ﺑﻦ‬ ‫ﻣﺤﻤﺪ‬ ‫اﻟﺮﻳﺤﺎن‬ ‫اﺑﻮ‬ ‫ﻫﻮ‬ : ( ‫379م‬ - 1048) = (‫263ﻫـ‬ - 440) ‫اﻟﺒﻴﺮوﻧﻲ‬ ✳
‫اج‬‫ﺮ‬‫ﻻﺳﺘﺨ‬ ‫ﻧﻈﺮﻳﺔ‬ ‫وﻟﻪ‬ ‫ﺑﺎﻓﻐﺎﻧﺴﺘﺎن‬ ‫ﻏﺰﻧﺔ‬ ‫ﻓﻲ‬ ‫وﺗﻮﻓﻲ‬ ‫ﺑﺨﻮارزم‬ ‫ﻛﺎث‬ ‫ﻓﻲ‬ ‫وﻟﺪ‬ ‫ﻓﺎرﺳﻲ‬ ‫اﺻﻞ‬ ‫ﻣﻦ‬ ‫ﻋﺮﺑﻲ‬
‫ﺣﻴﺚ‬ r = ‫ــــــــــــــــ‬ :‫ان‬ ‫ﻋﻠﻰ‬ ‫وﺗﻨﺺ‬ ‫اﻟﺒﻴﺮوﻧﻲ‬ ‫ﻗﺎﻧﻮن‬ ‫وﺗﺴﻤﻰ‬ ‫اﻻﺳﻄﺮﻻب‬ ‫ﻛﺘﺎﺑﻪ‬ ‫ﻓﻲ‬ ‫اﻻرض‬ ‫ﻣﺤﻴﻂ‬
‫اﻻﻓﻖ‬ ‫ﻋﻦ‬ ‫اﻻﻧﺤﺪار‬ ‫اوﻳﺔ‬‫ز‬ : x ، ‫اﻟﻤﺮﺻﻮد‬ ‫اﻻرﺗﻔﺎع‬ : b ، ‫ﺟﺒﻞ‬ ‫ﻗﻤﺔ‬ ‫ارﺗﻔﺎع‬ : a ، ‫اﻻرض‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ :r
‫اﺳﻤﺎﻋﻴﻞ‬ ‫ﺑﻦ‬ ‫ﻳﺤﻴﻰ‬ ‫ﻣﺤﻤﺪ‬ ‫ﺑﻦ‬ ‫ﻣﺤﻤﺪ‬ ‫ﻫﻮ‬ : (‫049م‬ - 988) = (‫823ﻫـ‬ - 388) : ‫اﻟﺒﻮزﺟﺎﻧﻲ‬ ✳
‫ﻣﻦ‬ ‫اول‬ ‫وﻫﻮ‬ . ‫ﺑﻐﺪاد‬ ‫اﻟﻰ‬ ‫أﻧﺘﻘﻞ‬ ‫م‬ 959 ‫ﻋﺎم‬ ‫وﻓﻲ‬ ‫ﺑﻮزﺟﺎن‬ ‫ﻣﺪﻳﻨﺔ‬ ‫ﻓﻲ‬ ‫وﻟﺪ‬ ، ‫اﻟﻮﻓﺎء‬ ‫أﺑﻮ‬ ‫اﻟﻌﺒﺎس‬ ‫ﺑﻦ‬
: ‫اﻻﺗﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ووﺿﻊ‬ ‫اﻟﺮﻳﺎﺿﻴﺔ‬ ‫اﻟﻤﺴﺎﺋﻞ‬ ‫ﺣﻞ‬ ‫ﻓﻲ‬ ‫واﺳﺘﻌﻤﺎﻟﻬﺎ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫وﺿﻊ‬
Sin x =2Sin ‫ــــــ‬ Cos ‫ــــــ‬
Sin(x+y)= Sin2
x -Sin2
x Siny + Sin2
y - Sin2
y Sin2
x
Tanx = ‫ـــــــــ‬ ، Cotx = ‫ـــــــــ‬
Secx = 1+Tan 2
x ، cecx = 1+ Cot2
x
‫ﺑﻦ‬ ‫ﻣﺤﻤﻮد‬ ‫ﺑﻦ‬ ‫ﻣﺴﻌﻮد‬ ‫ﺑﻦ‬ ‫ﺟﻤﺸﻴﺪ‬ ‫اﻟﺪﻳﻦ‬ ‫ﻏﻴﺎث‬ ‫ﻫﻮ‬ (‫2941م‬ ) = (‫ﻫـ‬ 899) ‫اﻟﻜﺎﺷﻲ‬ ✳
‫ﺗﻨﺴﺐ‬ ‫ﺣﻴﺚ‬ ‫ﻣﻤﺘﺎز‬ ‫رﻳﺎﺿﻲ‬ ‫ﻋﺎﻟﻢ‬ ‫ﻳﻌﺘﺒﺮ‬ ، ‫ﺳﻤﺮﻗﻨﺪ‬ ‫ﻓﻲ‬ ‫وﺗﻮﻓﻰ‬ ‫ان‬‫ﺮ‬‫ﺑﺎﻳ‬ ‫ﻛﺎﺷﺎن‬ ‫ﻣﺪﻳﻨﺔ‬ ‫ﻓﻲ‬ ‫وﻟﺪ‬ ‫اﻟﻜﺎﺷﻲ‬
‫ﺻﺤﻴﺤﺔ‬ 2 ‫اﻋﻄﻰ‬ ‫ﺣﻴﺚ‬ (‫اﻟﻤﺤﻴﻄﺔ‬ ‫)اﻟﺮﺳﺎﻟﺔ‬ ‫ﻛﺘﺎﺑﻪ‬ ‫ﻓﻲ‬ ‫وردت‬ ‫اﻟﺘﻲ‬ « π » ‫اﻟﺘﻘﺮﻳﺒﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫اﻟﻴﻪ‬
6.28318571795865 ، ً‫ﺎ‬‫ﻋﺸﺮﻳ‬ ً‫ﺎ‬‫رﻗﻤ‬ ‫ﻋﺸﺮ‬ ‫ﻟﺨﻤﺴﺔ‬
‫اﻟﺠﻴﺐ‬ ، ‫اﻟﻤﺴﺎﺣﺎت‬ ، ‫اﻟﻬﻨﺪﺳﺔ‬ ، ‫اﻟﺤﺴﺎب‬ : ‫ﻋﻠﻰ‬ ‫ﺣﻮى‬ ‫واﻟﺬي‬ «‫اﻟﺤﺴﺎب‬ ‫»ﻣﻔﺘﺎح‬ ‫ﻣﺆﻟﻔﺎﺗﻪ‬ ‫وﻣﻦ‬
(.... ‫وﻏﻴﺮﻫﺎ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﺟﻴﺐ‬ ‫اج‬‫ﺮ‬‫اﺳﺘﺨ‬ ،‫واﻟﻮﺗﺮ‬
- 1617) ‫ﻧﺎﺑﻴﺮ‬ ‫ﺟﻮن‬ ‫اﻻﺳﻜﺘﻠﻨﺪي‬ ‫اﻟﻌﺎﻟﻢ‬ ‫ﻳﺪ‬ ‫ﻋﻠﻰ‬ ‫ﻋﺸﺮ‬ ‫اﻟﺴﺎﺑﻊ‬ ‫اﻟﻘﺮن‬ ‫ﻓﻲ‬ ‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﻋﻠﻢ‬ ‫ﺗﻄﻮر‬ ‫وﻗﺪ‬
‫ﻣﻦ‬ ‫وﻛﺜﻴﺮ‬ ‫واﻟﻔﻴﺰﻳﺎء‬ ‫اﻓﻴﺔ‬‫ﺮ‬‫واﻟﺠﻐ‬ ‫واﻟﻤﺴﺎﺣﺔ‬ ‫اﻟﻤﻼﺣﺔ‬ ‫ﻓﻲ‬ ‫ﻛﺜﻴﺮة‬ ‫اﺳﺘﺨﺪاﻣﺎت‬ ‫اﻟﻤﺜﻠﺜﺎت‬ ‫وﻟﻌﻠﻢ‬ (‫م‬ 1550
. ‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﻟﻤﻮﺿﻮع‬ ‫اﻻﺳﺎﺳﻴﺔ‬ ‫اﻟﻤﺒﺎدئ‬ ‫ﺳﻨﻌﻄﻲ‬ ‫اﻟﻔﺼﻞ‬ ‫ﻫﺬا‬ ‫وﻓﻲ‬ ‫اﻟﻬﻨﺪﺳﺔ‬ ‫ﻓﺮوع‬
b Cos x
a - Cos x
x
2
Sin x
Cos x
Cos x
Sin x
x
2
، 2Sin2
‫ــــــ‬ = Cos x1-x
2
2π =
π

60
� ‫اﻟﻘﻴﺎﺳﻲ‬ ‫ﺑﺎﻟﻮﺿﻊ‬ ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ [4 � 1 �
(4 - 1 ) ‫ﺗﻌﺮﻳﻒ‬
‫ﻫﻲ‬ ‫ﻣﺸﺘﺮﻛﺔ‬ ‫ﺑﺪاﻳﺔ‬ ‫ﻧﻘﻄﺔ‬ B C ، B A ‫ﻟﻠﺸﻌﺎﻋﻴﻦ‬ ‫ﻛﺎن‬ ‫اذا‬ : Directed Angle ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
‫وﺿﻠﻌﻬﺎ‬ B A ‫اﻻﺑﺘﺪاﺋﻲ‬ ‫ﺿﻠﻌﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻳﺴﻤﻰ‬ ( B A, B C ) ‫اﻟﻤﺮﺗﺐ‬ ‫اﻟﺰوج‬ ‫ﻓﺎن‬ B
. A B C ‫أو‬ ( B A ، B C ) ‫اﻟﻄﺮﻳﻘﺘﻴﻦ‬ ‫ﺑﺎﺣﺪى‬ ‫وﺗﻜﺘﺐ‬ B ‫اﻟﻨﻘﻄﺔ‬ ‫ورأﺳﻬﺎ‬ B C ‫اﻟﻨﻬﺎﺋﻲ‬
( 4 - 2 ) ‫ﺗﻌﺮﻳﻒ‬
‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫ﻣﺘﻌﺎﻣﺪ‬ ‫اﺣﺪاﺛﻲ‬ ‫ﻧﻈﺎم‬ ‫ﻟﺪﻳﻨﺎ‬ ‫ﻛﺎن‬ ‫اذا‬ : ‫اﻟﻘﻴﺎﺳﻲ‬ ‫ﺑﺎﻟﻮﺿﻊ‬ ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
‫ﻧﻘﻄﺔ‬ ‫ﻓﻲ‬ ‫رأﺳﻬﺎ‬ ‫وﻗﻊ‬ ‫اذا‬ ‫ﻗﻴﺎﺳﻲ‬ ‫وﺿﻊ‬ ‫ﻓﻲ‬ ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ان‬ ‫ﻓﻴﻘﺎل‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫ﻣﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ز‬‫و‬
. ( 4 - 1 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻟﺠﺰء‬ ‫ﻋﻠﻰ‬ ‫اﻻﺑﺘﺪاﺋﻲ‬ ‫ﺿﻠﻌﻬﺎ‬ ‫واﻧﻄﺒﻖ‬ ‫اﻷﺻﻞ‬
( 4 - 2 ) ‫اﻟﺸﻜﻞ‬ ( 4 - 1 ) ‫اﻟﺸﻜﻞ‬
>
←
←←
A
C
B
•
•
•C
•
•
A
B
O
Y
XX
‫اﻟﻤﻮﺟﺒﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
( ‫اﻟﺴﺎﻋﺔ‬ ‫ﻋﻘﺮب‬ ‫ان‬‫ر‬‫دو‬ ‫ﻋﻜﺲ‬ )
Y
‫اﻟﺴﺎﻟﺒﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
( ‫اﻟﺴﺎﻋﺔ‬ ‫ﻋﻘﺮب‬ ‫ان‬‫ر‬‫دو‬ ‫ﻣﻊ‬ )
←
← ←←
←←
XX

61
� ‫ﻟﻠﺰواﻳﺎ‬ ‫اﻟﺪاﺋﺮي‬ ‫واﻟﻘﻴﺎس‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻘﻴﺎس‬ [ 4 � 2 �
:‫اﻧﻪ‬ ‫اﻟﻤﺘﻮﺳﻄﺔ‬ ‫اﻟﻤﺮﺣﻠﺔ‬ ‫ﻣﻦ‬ ‫ﺗﻌﻠﻤﻨﺎ‬ : Degree Measure ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻘﻴﺎس‬
‫ﻣﻨﻬﺎ‬ ‫ﻗﻮس‬ ‫ﻛﻞ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬ ً‫ﺎ‬‫ﻗﻮﺳ‬ 360 ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ‫ﻓﺎﻧﻨﺎ‬ ً‫ﺎ‬‫ﻣﺘﺴﺎوﻳ‬ ً‫ﺎ‬‫ﻗﺴﻤ‬ 360 ‫ﻋﻠﻰ‬ ‫داﺋﺮة‬ ‫ﻗﺴﻤﻨﺎ‬ ‫اذا‬
1˚ ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫ﻓﻲ‬ ‫درﺟﺔ‬ ‫ﻳﺴﻤﻰ‬ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ ‫ﻣﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻳﻘﺎﺑﻞ‬
360´´ = ‫ﺛﺎﻧﻴﺔ‬ 60 = 1´ ، 60´ = ‫دﻗﻴﻘﺔ‬ 60 = 1˚ : ‫ان‬ ‫ﻛﻤﺎ‬
‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻴﺎس‬ ‫ﻳﺴﻤﻰ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻟﻘﻴﺎس‬ ‫اﺧﺮ‬ ‫ﻧﻈﺎم‬ ‫ﻳﻮﺟﺪ‬ : Radian Measure ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻴﺎس‬
.‫ﻟﻠﺰواﻳﺎ‬
(4 - 3) ‫ﺗﻌﺮﻳﻒ‬
‫وﺿﻊ‬ ‫اﻟﺘﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ ‫ﻗﻴﺎس‬ ‫وﻫﻲ‬ ‫اﻟﻘﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻫﻲ‬ ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻗﻴﺎس‬ ‫وﺣﺪة‬
. ‫اﻟﺪاﺋﺮة‬ ‫ﺗﻠﻚ‬ ‫ﻗﻄﺮ‬ ‫ﻟﻨﺼﻒ‬ ٍ‫و‬‫ﻣﺴﺎ‬ ‫ﻃﻮﻟﻪ‬ Arc ‫ﻗﻮس‬ ‫وﻗﺎﺑﻠﻬﺎ‬ ‫داﺋﺮة‬ ‫ﻣﺮﻛﺰ‬ ‫ﻓﻲ‬ ‫رأﺳﻬﺎ‬
‫وﺣﺪة‬ (L) ‫ﻳﺴﺎوي‬ A O B ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ ‫اﻟﻤﻘﺎﺑﻞ‬ ‫اﻟﻘﻮس‬ ‫ﻃﻮل‬ ‫ان‬ ‫ﻓﺮﺿﻨﺎ‬ ‫اذا‬ (4 - 3 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻔﻲ‬
A O B ‫ﻓﺎن‬ L = r ‫وﻛﺎن‬ ‫ﻃﻮل‬ ‫وﺣﺪة‬ r = ‫اﻟﺪاﺋﺮة‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬
. ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ 1 = ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬
A O B ‫ﻓﺎن‬ (4 - 4) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ L = 2r ‫ﻛﺎن‬ ‫واذا‬
. ‫اﻟﻘﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻣﻦ‬ 2 = ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬
: ‫ﻫﻮ‬ r ‫ﻗﻄﺮﻫﺎ‬ ‫ﻧﺼﻒ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻗﻮس‬ ‫ﻃﻮل‬ ‫ان‬ ‫ﻳﻨﺘﺞ‬ (4 - 3) ‫ﺗﻌﺮﻳﻒ‬ ‫وﻣﻦ‬
.‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ً‫ا‬‫ر‬‫ﻣﻘﺪ‬ ‫اﻟﻘﻮس‬ ‫ﻟﺬﻟﻚ‬ ‫اﻟﻤﻘﺎﺑﻠﺔ‬ ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﻴﺎس‬ | Q | ‫ﺣﻴﺚ‬
| Q | = ‫ــــــ‬ = ‫ــــــــــــــــــــ‬
•
A
B
L
r •o
r
•
B
L
r
•
o
r
(4-3)‫اﻟﺸﻜﻞ‬
(4-4)‫اﻟﺸﻜﻞ‬
L
r
‫اﻟﻘﻮس‬ ‫ﻃﻮل‬
‫اﻟﻘﻄﺮ‬ ‫ﻧﺼﻒ‬
m >
m >
A
‫أو‬
L= Q . r

62
� ‫ﻟﻠﺰواﻳﺎ‬ ‫واﻟﺪاﺋﺮي‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻘﻴﺎس‬ ‫ﺑﻴﻦ‬ ‫اﻟﻌﻼﻗﺔ‬ [ 4 � 3 �
2πr = ‫اﻟﺪاﺋﺮة‬ ‫ﻣﺤﻴﻂ‬ ‫ان‬ ً‫ﺎ‬‫ﺳﺎﺑﻘ‬ ‫ﺗﻌﻠﻤﻨﺎ‬
‫ــــــ‬ = ‫ـــــــــ‬ ‫ان‬ ‫وﺑﻤﺎ‬
360˚ = ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ 2π ‫ان‬ ‫وﺑﻤﺎ‬
180˚ = ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ π ∴
‫ــــــ‬ = ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ 1 ⇐
. ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ ‫ــــــ‬ = 1˚ ⇐
‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ Q = ‫ﻣﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫ﻛﺎن‬ ‫اذا‬ ( ‫أ‬ : ‫ﻋﺎﻣﺔ‬ ‫ﺑﺼﻮرة‬
‫ــــــــــــ‬ = ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ Q ‫ﻓﺈن‬
: ‫ﻓﺈن‬ D˚ = ‫ﻣﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫ﻛﺎن‬ ‫اذا‬ ( ‫ب‬
: ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ ( ‫ــــــ‬ × ) = D˚
‫ــــــ‬ = ‫ــــــ‬ : ‫ان‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫وﻣﻨﻪ‬
. ‫وﺑﺎﻟﻌﻜﺲ‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻰ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫ﻣﻦ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫ﻟﺘﺤﻮﻳﻞ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﻫﺬه‬ ‫وﺗﺴﺘﺨﺪم‬
L
r
2πr
r
180˚
π
π
˚180
D˚× π
180˚
180˚
π
Q
D˚
π
180˚
| Q | = = 2 π
Q

63
� � 1 ‫ﻣﺜﺎل‬
‫ﻗﻄﺮﻫﺎ‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ 10cm ‫ﻃﻮﻟﻪ‬ ً‫ﺎ‬‫ﻗﻮﺳ‬ ‫ﺗﻘﺎﺑﻞ‬ ‫ﻗﻴﺎﺳﻲ‬ ‫وﺿﻊ‬ ‫ﻓﻲ‬ A O B ‫ﻛﺎﻧﺖ‬ ‫اذا‬
. 12cm
: ‫ﺣﻴﺚ‬ A O B ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫اﺣﺴﺐ‬ ( ‫أ‬
.‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻫﻮ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻣﺮﻛﺰ‬ ‫ان‬ ً‫ﺎ‬‫ﻋﻠﻤ‬
: ‫ﺣﻴﺚ‬ m A O B ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫اﺣﺴﺐ‬ (‫ب‬
L = 10 cm ، r = 12 cm
| Q | = ‫ــــــ‬ = ‫ــــــ‬ = ‫ــــــ‬ = 0. 833 ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ ∴ ( ‫أ‬
: ‫وﻳﻜﻮن‬ ً‫ﺎ‬‫ﺳﺎﻟﺒ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﻴﺎس‬ ‫ﻳﻜﻮن‬ ‫اﻟﺤﺎﻟﺔ‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ (‫ب‬
| Q | = ‫ــــــ‬ = ‫ــــــ‬ = ‫ــــــ‬ = 0. 833
.( ‫ﺳﺎﻟﺒﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻻن‬ )‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ Q = - 0. 833 ∴
� � 2 ‫ﻣﺜﺎل‬
‫؟‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﻓﻤﺎ‬ ‫ــــــ‬ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫وﻛﺎن‬ ‫اﻟﻘﻴﺎﺳﻲ‬ ‫وﺿﻌﻬﺎ‬ ‫ﻓﻲ‬ A O B ‫ﻛﺎﻧﺖ‬ ‫اذا‬
‫ــــــ‬ = ‫ــــــ‬
‫ـــــــــ‬ = ‫ــــــ‬ ⇒ D˚ = 180 ˚ × ‫ــــــ‬ = ˚
L
r
10
12
5
6
10
12
5
6
3π
4
Q
D˚
π
180˚
‫ــــــ‬3π
4
D˚
3
4
←
<
←
<m
2π m A O B 0 ≤ < ≤
< ←
0 m A O B -2π≤ < <
L
r
←
<
135π
180˚

64
1
4
� � 3 ‫ﻣﺜﺎل‬
. ‫اﻟﺪاﺋﺮي‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﻰ‬ 45˚ ( ‫أ‬ : ‫ﺣﻮل‬
. ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﻰ‬ 2.6 ( ‫ب‬
. ‫ﻗﻄﺮﻳﺔ‬ ‫اﻟﻨﺼﻒ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻣﻦ‬ ‫ــــــ‬ = Q ⇐ ‫ــــــ‬ = ‫ــــــ‬ ⇐ ‫ــــــ‬ = ‫∵ــــــ‬ ( ‫أ‬
468˚ = 2.6 × 180˚ = ˚ ⇐ ‫ــــــــ‬ = ‫ــــــ‬ ⇐ ‫ــــــ‬ = ‫ ــــــ‬ ( ‫ب‬
� � 4 ‫ﻣﺜﺎل‬
‫؟‬ 9cm ‫داﺋﺮﺗﻬﺎ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬ ‫ﻛﺎن‬ ‫اذا‬ ‫ﺗﻘﺎﺑﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻘﻮس‬ ‫ﻃﻮل‬ ‫ﻓﻤﺎ‬ 60˚ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﻣﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ز‬
. ‫ﻗﻄﺮﻳﺔ‬ ‫اﻟﻨﺼﻒ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻣﻦ‬ ‫ــــــ‬ = Q ⇐ ‫ــــــ‬ = ‫ــــــ‬ ⇐ ‫ــــــ‬ = ‫∵ــــــ‬
|Q|= ‫∵ ــــــ‬
� � 5 ‫ﻣﺜﺎل‬
‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﻣﻘﺪار‬ ‫02ﻓﻤﺎ‬ ‫داﺋﺮﺗﻬﺎ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 21 ‫ــــــ‬ ‫ﻗﻮﺳﻬﺎ‬ ‫ﻃﻮل‬ ‫ﻣﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ز‬
‫اﻟﺴﺘﻴﻨﻲ؟‬
Q
D˚
π
180˚
π
180˚
Q
45 ˚
π
4
Q
D˚
π
180˚
π
180˚
Q
D˚
π
180˚
Q
60˚
π
180˚
1
3
L
r
π
D
π
3π = 3 × 3.142
= 9.426 cm
L
9
‫ــــــ‬ = ‫ــــــ‬π
3 ⇒ L =
π
cm cm
2.6
D˚

65
L
r
Q
D˚
π
180˚
‫ــــــ‬17
16
D
π
180˚
17
16
A + B = 90˚ . . . . . . . 1
A - B = 25.2˚ . . . . . . 2
2 A = 115.2
∴ A = 57.6˚
B = 32.4˚
7
22
π
180˚
Q
D˚
π
180˚
D˚ =
0.44
D˚
0.44 × 180
π
0.44 × 180
3.14
= =25.2˚
. ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫زاوﻳﺔ‬ ‫ــــــ‬ = ‫ـــــــــــــ‬ =|Q| ⇐ ‫ــــــ‬ =|Q|
‫ــــــ‬ = ‫ــــــ‬ ⇐ ‫ــــــ‬ = ‫ــــــ‬ ‫ﺛﻢ‬
D˚= × 180˚ × = 60. 85˚
� � 6 ‫ﻣﺜﺎل‬
‫ﻣﻨﻬﺎ‬ ‫ﻛﻞ‬ ‫ﻗﻴﺎس‬ ‫ﻓﻤﺎ‬ ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ 0.44 ‫اﻟﺤﺎدﺗﻴﻦ‬ ‫اوﻳﺘﻴﻪ‬‫ز‬ ‫ﺑﻴﻦ‬ ‫اﻟﻔﺮق‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ ‫ﻓﻲ‬
‫؟‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬
‫ــــــ‬ = ‫ــــــ‬ ⇐ ‫ـــــ‬ = ‫∵ــــــ‬
∴
A ، B ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﻗﻴﺎﺳﻬﻤﺎ‬ ‫اﻟﺤﺎدﺗﻴﻦ‬ ‫اوﻳﺘﻴﻦ‬‫ﺰ‬‫اﻟ‬ ‫ان‬ ‫ﻧﻔﺮض‬
D˚
17
16
21 ‫ــــــ‬1
4
20
Q
Do
=
π
180o
: ‫ﻫﻲ‬ Q ‫ﻗﻄﺮي‬ ‫اﻟﻨﺼﻒ‬ ‫واﻟﻨﻈﺎم‬ D0
‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻨﻈﺎم‬ ‫ﺑﻴﻦ‬ ‫اﻟﻌﻼﻗﺔ‬
Q =
L
r
:‫ﻫﻲ‬ r ‫داﺋﺮﺗﻬﻢ‬ ‫ﻗﻄﺮ‬ ‫وﻧﺼﻒ‬ L ‫اﻟﻘﻮس‬ ‫وﻃﻮل‬ Q ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻌﻼﻗﺔ‬

66
( 4 - 1 ) ‫تمرينات‬
/ 1‫س‬
: ‫اﻻﺗﻴﺔ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻗﻴﺎس‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﻰ‬ ‫ﺣﻮل‬
300 ˚ ، 120 ˚ ، 30 ˚
/ 2‫س‬
: ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﻰ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻘﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻣﻦ‬ ً‫ﻼ‬‫ﻛ‬ ‫ﺣﻮل‬
‫ـــــــ‬ ، ‫ـــــــ‬ ، ‫ـــــــ‬
/ 3‫س‬
‫ﻧﺼﻒ‬ ‫ﺟﺪ‬ 25 ‫ﻃﻮﻟﻪ‬ً‫ﺎ‬‫ﻗﻮﺳ‬ ‫ﺗﻘﺎﺑﻞ‬ ‫اﻟﻘﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻣﻦ‬ ‫ــــــ‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ ‫ﻣﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻗﻴﺎس‬
30 / ‫ج‬ . ‫اﻟﺪاﺋﺮة‬ ‫ﺗﻠﻚ‬ ‫ﻗﻄﺮ‬
/ 4‫س‬
‫؟‬ 8cm ‫ﻗﻄﺮﻫﺎ‬ ‫ﻧﺼﻒ‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ 135˚ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﻣﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﻤﻘﺎﺑﻞ‬ ‫اﻟﻘﻮس‬ ‫ﻣﺎﻃﻮل‬
18.857 / ‫ج‬
/ 5‫س‬
‫اوﻳﺘﻴﻦ‬‫ﺰ‬‫اﻟ‬ ‫ﻫﺎﺗﻴﻦ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬ 9˚ ‫ﻳﺴﺎوي‬ ‫وﻓﺮﻗﻬﻤﺎ‬ ‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ ‫ــــــ‬ ‫ﻣﺠﻤﻮﻋﻬﻤﺎ‬ ‫اوﻳﺘﺎن‬‫ز‬
18˚ ، 27˚ / ‫ج‬ ‫؟‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬
/ 6‫س‬
‫؟‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ ‫ـــــــ‬ = ‫ﻗﻴﺎﺳﻬﺎ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﻘﻴﺎﺳﻲ‬ ‫وﺿﻌﻬﺎ‬ ‫ﻓﻲ‬ ‫ارﺳﻢ‬
3 π
5
5 π
6
1
3
5
6
π
4
←5 π
4 AOB>
cm
cm
cm

67
‫اﻻﺳﺎﺳﻴﺔ‬ ‫اﻟﻌﻼﻗﺎت‬ ‫وﺑﻌﺾ‬ ‫ﺣﺎدة‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺐ‬ [4 � 4 �
C ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻳﻤﺜﻞ‬ (4 - 5) ‫اﻟﺸﻜﻞ‬
‫وﻟﻴﻜﻦ‬
( 4 - 5 ) ‫اﻟﺸﻜﻞ‬
(4 - 4) ‫ﺗﻌﺮﻳﻒ‬
: ‫ﻳﻠﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻻﺗﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫ﻳﻤﺜﻞ‬ ‫اﻟﺬي‬ ‫اﻟﻌﺪد‬ ‫ﻧﺴﻤﻲ‬
( Q ) ‫اﻟﺤﺎدة‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ( Sine ) ‫ﺟﻴﺐ‬ ‫ﺗﺪﻋﻰ‬ ‫ــــــ‬ ‫اﻟﻨﺴﺒﺔ‬ .1
(Q) ‫اﻟﺤﺎدة‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ( Cosine ) ‫ﺗﻤﺎم‬ ‫ﺟﻴﺐ‬ ‫ﺗﺪﻋﻰ‬ ‫ــــــــ‬ ‫اﻟﻨﺴﺒﺔ‬ .2
‫وﺗﻜﺘﺐ‬
(Q) ‫اﻟﺤﺎدة‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ( Tangent ) ‫ﻇﻞ‬ ‫ﻓﺘﺪﻋﻰ‬ ‫ـــــــــ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫اﻣﺎ‬ .3
‫وﺗﻜﺘﺐ‬
‫ﺮ‬‫ﺗ‬‫ﻮ‬‫ﻟ‬‫ا‬
A
B C
‫اﻟﻤﻘﺎﺑﻞ‬
‫اﻟﻤﺠﺎور‬
Hypotenuse
Opposite
Adjacent
Q
AC
AB
OPP.
HYP.
AC
BC
ABC
m < ABC = Q
‫وﺗﻜﺘﺐ‬Sin Q = ‫ـــــــ‬ = ‫ـــــــــ‬AC
AB
BC
AB
ADJ.
HYP.
Cos Q = ‫ـــــــ‬ = ‫ـــــــــ‬BC
AB
OPP.
ADJ.
tan Q = ‫ـــــــ‬ = ‫ـــــــــ‬AC
BC

68
AB
AB
(AC)2
+(BC)2
= (AB)2
(AB)2
2
Sin2
Q + Cos2
Q = 1
tan Q = ‫ـــــــ‬AC
BC
tan Q = ‫ـــــــــــ‬
‫ـــــــ‬BC
AB
‫ـــــــ‬AC
AB
tan Q = ‫ـــــــــــ‬Sin Q
Cos Q
AC
AB
BC
AB
(‫ﻓﻴﺜﺎﻏﻮرس‬‫ﻣﺒﺮﻫﻨﺔ‬) :(4 - 5 ) ‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬
: ‫ﻳﻨﺘﺞ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫وﺑﻘﺴﻤﺔ‬
‫ــــــــ‬ + ‫ــــــــ‬ = ‫ــــــــ‬
( 4 - 4 ) ‫ﺗﻌﺮﻳﻒ‬ ‫وﻣﻦ‬
: ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬
: ً‫ﺎ‬‫آﻳﻀ‬ (4-4) ‫ﺗﻌﺮﻳﻒ‬ ‫ﻣﻦ‬
: ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ( AB ) ‫ﻋﻠﻰ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫وﺑﻘﺴﻤﺔ‬
∴
2 2

69
� Trigonometric Ratio ‫ﺧﺎﺻﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ [4 � 5 �
45 ˚ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫اوﻳﺔ‬‫ز‬ ( 1
(45˚) ‫اﻻﺧﺮى‬ ‫ﻓﺘﻜﻮن‬ (45˚) ‫ﻗﻴﺎﺳﻬﺎ‬ ‫زواﻳﺎه‬ ‫واﺣﺪى‬ . B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ABC ‫اﻟﻤﺜﻠﺚ‬ ‫ﻧﺮﺳﻢ‬
ً‫ﺎ‬‫اﻳﻀ‬
AB = BC = L ∴
:‫ﻓﻴﺜﺎﻏﻮرس‬
(AC)2
= L2
+L2
= 2L2
∴
60 ˚ ،30 ˚ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫اوﻳﺔ‬‫ز‬ ( 2
2L = ‫ﺿﻠﻌﻪ‬ ‫ﻃﻮل‬ ‫اﻷﺿﻼع‬ ‫ﻣﺘﺴﺎوي‬ ً‫ﺎ‬‫ﻣﺜﻠﺜ‬ ‫ﻧﺮﺳﻢ‬
60˚ = ‫ﻣﻨﻬﺎ‬ ‫وﻛﻞ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬ ‫زواﻳﺎه‬ ‫ﻗﻴﺎﺳﺎت‬ ‫ﻓﻴﻜﻮن‬
‫اﻟﻤﺠﺎور‬ ‫اﻟﺸﻜﻞ‬ ‫ﻻﺣﻆ‬ AD⊥BC ‫ﻧﺮﺳﻢ‬
‫وﺣﺪة‬ CD = DB = L ∴
m BAD = 30 ˚ ‫ن‬ ‫وا‬
= ‫ان‬ ‫ﻧﺠﺪ‬ ‫ﻓﻴﺜﺎﻏﻮرس‬ ‫ﻣﺒﺮﻫﻨﺔ‬ ‫ﺑﺎﺳﺘﺨﺪام‬
∀
A
CB ˚60L
2L
L
L˚60
˚30 ˚30
2L
D
(AC)2
= (AB)2
+ (BC)2
AC = 2 L
Sin 45˚ = ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ Sin 45˚= ‫ـــــــ‬L
2L
1
2
1
2
Cos 45˚= ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ Cos 45˚= ‫ـــــــ‬L
2L
1
2
1
2
tan 45˚= ‫ـــــــ‬ = ⇒ tan 45˚ = 1L
L
1
AD 3 L
Sin 30˚ = ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ Sin 30˚= ‫ـــــــ‬L
2L
1
2
1
2
3L
2L
sin 60˚ = ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ sin 60˚= ‫ـــــــ‬3
2
3L
⇒L 1
A
C B
45˚
L
2
45˚
L
L
3
2

70
: ‫ان‬ ‫ﻻﺣﻆ‬
‫ﻛﺬﻟﻚ‬
.‫وﺑﺎﻟﻌﻜﺲ‬ ‫اﻻﺧﺮى‬ ‫ﺗﻤﺎم‬ ‫ﺟﻴﺐ‬ ‫ﻳﺴﺎوي‬ ‫اﺣﺪﻫﻤﺎ‬ ‫ﺟﻴﺐ‬ ‫ان‬ ‫اي‬
‫ﻣﺘﻤﻤﺘﻬﺎ‬ ‫ﻗﻴﺎس‬ ‫ﻓﺎن‬ ‫ﺣﺎدة‬ ‫اوﻳﺔ‬‫ز‬ Q ‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫ﻋﺎﻣﺔ‬ ‫وﺑﺼﻮرة‬
: ‫وﻳﻜﻮن‬ ( ) ‫ﻫﻮ‬
‫ﻣﺘﺘﺎﻣﺘﺎن‬60˚,30˚:‫اوﻳﺘﺎن‬‫ﺰ‬‫اﻟ‬
‫ﻷن‬
60˚ + 300
= 90˚
Cos 30˚= ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ Cos 30˚= ‫ـــــــ‬3L
2L
3
2
3
2
Cos 60˚= ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ Cos 60˚= ‫ـــــــ‬L
2L
1
2
1
2
tan 30˚ = ‫ـــــــ‬ = ‫ـــــــ‬ ⇒ tan 30˚= ‫ـــــــ‬L
3L
1
3
1
3
tan 60˚ = ‫ـــــــ‬ = ⇒ tan 60˚=3L
L
33
sin 30˚ = 1
2
cos 60˚ = ‫ـــــــ‬
3
2
cos 30˚ = sin 60˚ = ‫ـــــــ‬
sin (90˚ -Q ) = cos Q
cos (90˚ -Q ) = sin Q
90˚ - Q
* Sin2
Q + cos2
Q = 1 , tan Q =
sin
cos Q
* sin (90o
− Q) = cos Q , cos (90o
− Q) = sinQ
* sin30o
= cos 60o
=
1
2
, cos 30o
= sin 60o
=
3
2
* sin 45o
= cos 45o
=
1
2
* sin Q = , cos Q = , tan Q =
‫اﻟﻮﺗﺮ‬ ‫اﻟﻮﺗﺮ‬
‫اﻟﻤﺠﺎور‬ ‫اﻟﻤﻘﺎﺑﻞ‬
sinQ
cosQ

71
� ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫واﻟﻨﻘﻄﺔ‬ ‫اﻟﻮﺣﺪة‬ ‫داﺋﺮة‬ [ 4 � 6 �
(4 - 5) ‫ﺗﻌﺮﻳﻒ‬
. ‫واﺣﺪة‬ ‫ﻃﻮل‬ ‫وﺣﺪة‬ ‫ﻳﺴﺎوي‬ ‫ﻗﻄــﺮﻫــــﺎ‬ ‫وﻧﺼﻒ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﺮﻛﺰﻫﺎ‬ ‫داﺋﺮة‬ ‫ﻫﻲ‬ : ‫اﻟﻮﺣﺪة‬ ‫داﺋﺮة‬
، m B O A = Q ‫اﻟﺸﻜــﻞ‬ ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﻘﻄﺔ‬
‫اﻟﻀﻠﻊ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬ B ، ‫اﻟﻘﻴﺎﺳﻲ‬ ‫اﻟﻮﺿﻊ‬ ‫ﻓﻲ‬ ‫ﻣﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ز‬
‫ان‬ ‫ﻧﻔﺮض‬ ‫اﻟﻮﺣــﺪة‬ ‫داﺋﺮة‬ ‫ﻣﻊ‬ ‫اﻟﻨﻬﺎﺋﻲ‬
sin Q = ‫ــــــ‬ ⇒ sin
cos Q = ‫ــــــ‬ ⇒ cos
∴
(4 - 6) ‫ﺗﻌﺮﻳﻒ‬
‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻫﻲ‬ ‫اﻟﻘﻴﺎﺳﻲ‬ ‫اﻟﻮﺿﻊ‬ ‫ﻓﻲ‬ ‫اﻟﻤﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ Trigonometric Point ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﻘﻄﺔ‬
. ‫اﻟﻮﺣﺪة‬ ‫داﺋﺮة‬ ‫ﻣﻊ‬ ‫ﻟﻠﺰواﻳﺔ‬ ‫اﻟﻨﻬﺎﺋﻲ‬ ‫اﻟﻀﻠﻊ‬
∀
y
1
A
y
xO
y
B(x,y )
x
←
←
‫اﻟﻮﺣﺪة‬ ‫داﺋﺮة‬ ‫ﺑﺎﺳﺘﺨﺪام‬
‫اﻟﻤﺴﺘﻮي‬ ‫ﻋﻠﻰ‬ ‫واﻻﻧﻌﻜﺎس‬
‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺐ‬ ‫إﻳﺠﺎد‬ ‫ﻳﻤﻜﻦ‬
: ‫اﻵﺗﻴﺔ‬
sin (180˚-Q ) = sin Q
cos (180˚-Q ) = -cos Q
tan (180˚-Q ) = -tan Q
x
1
B(x,y) = (cosQ,sin Q)
B(x,y) O B
Q
Q= Y
Q= X

72
←
Q ‫ﻣﻮﺟﻬﺔ‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻟﻜﻞ‬ ‫أن‬ ‫ﻳﺘﻀﺢ‬ ‫ﺳﺒﻖ‬ ‫ﻣﻤﺎ‬ AOB ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ ‫ﻣﺜﻠﺜﻴﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻫﻲ‬ B ‫ﻧﻘﻄﺔ‬ ‫أن‬ ‫ﻻﺣﻆ‬
. x = ، = ‫ﻳﻜﻮن‬ (x , y) ‫ﻣﺜﻠﺜﻴﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫اﻟﻘﻴﺎﺳﻲ‬ ‫اﻟﻮﺿﻊ‬ ‫ﻓﻲ‬
� � 7 ‫ﻣﺜﺎل‬
‫أن‬ ‫ﻋﻠﻤﺖ‬ ‫إذا‬ ، ، tan Q ‫ﺟﺪ‬
‫ﻓﻲ‬ ‫وﻛﻤﺎ‬ . ‫اﻻﺣﺪاﺛﻴﻴﻦ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫أﺣﺪ‬ ‫ﻋﻠﻰ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﻨﻬﺎﺋﻲ‬ ‫اﻟﻀﻠﻊ‬ ‫ﻳﻘﻊ‬ ˚180 ،˚90 ، ˚0 ‫ان‬ ‫ﻧﻌﻠﻢ‬
: ‫ﻓﺎن‬ ( 4 - 6 ) ‫اﻟﺸﻜﻞ‬
∴tan 0˚ = ‫ـــــــ‬ = ‫ــــــــ‬ ⇒ tan˚0 = 0
(cos 90˚ , sin 90˚ ) = (0,1 ) ✳
⇒ cos 90˚= 0 ، sin 90˚=1
‫ﻣﻌﺮف‬ ‫ﻏﻴﺮ‬ tan90˚ = ‫ــــــــ‬ ‫ﻟﻜﻦ‬
(cos 180˚ , sin 180˚) = ( -1, 0) ✳
⇒ cos 180˚= -1 ، sin 180˚= 0
⇒ tan 180˚= 0
(4 - 6) ‫اﻟﺸﻜﻞ‬
sin˚0
cos˚0
0
1
sin90˚
cos90˚
(
y sin Qcos Q
cos Qsin QQ = 0˚ ، 90˚، 180˚
(cos 0 , sin 0) = (1 , 0) ⇒ cos 0˚ = 1
sin 0˚ = 0
C (-1,0)
Y
(cos0 , sin0)
=(1,0)
X
•
•×
×
×
B (cos90˚, sin90˚)
=(0,1)
O
A
=0

73
� ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﺘﻄﺒﻴﻘﺎت‬ [4 � 7 �
:‫واﻻﻧﺨﻔﺎض‬ ‫اﻻرﺗﻔﺎع‬ ‫اوﻳﺘﺎ‬‫ز‬ [4-7-1]
‫وﻗﻒ‬ ‫ﻓﺈذا‬ .‫ﺑﻬﺎ‬ ‫اﻫﺎ‬‫ﺮ‬‫ﻧ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺰواﻳﺎ‬ ‫ﻗﻴﺎس‬ ‫ﻣﻦ‬ ‫ﻧﺘﻤﻜﻦ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫واﻻﺑﻌﺎد‬ ‫اﻻرﺗﻔﺎﻋﺎت‬ ‫ﺣﺴﺎب‬ ‫ﻣﻦ‬ ‫ﻧﺘﻤﻜﻦ‬
‫اﻟﻮاﺻﻞ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺤﺎﺻﻠﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻓﺈن‬ A ‫اﻓﻖ‬ ‫ﻓﻮق‬ ‫ﺗﻘﻊ‬ C ‫ﻧﻘﻄﺔ‬ ‫اﻟﻰ‬ ‫وﻧﻈﺮ‬ A ‫ﻧﻘﻄﺔ‬ ‫ﻓﻲ‬ ‫اﺻﺪ‬‫ر‬
‫ﺑﺎﻟﻨﺴﺒﺔ‬Angle of Elevation C ‫إرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬) ، ‫ﺗﺪﻋﻰ‬ A ‫أﻓﻖ‬ ‫وﺑﻴﻦ‬ C ‫ﻧﻘﻄﺔ‬ ‫اﻟﻰ‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﻋﻴﻦ‬ ‫ﻣﻦ‬
.(4 - 7) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ CAB ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ً‫ﻼ‬‫ﻣﺜ‬ (A ‫اﻟﻰ‬
‫اﻟﻰ‬ ‫وﻧﻈﺮ‬ C ‫ﻓﻲ‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﻋﻴﻦ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫أﻣﺎ‬
، ‫اﻟﻜﺎﺋﻨﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻓﺈن‬ ، C ‫أﻓﻖ‬ ‫ﺗﺤﺖ‬ ‫اﻟﺘﻲ‬ A
‫إﻟﻰ‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﻋﻴﻦ‬ ‫ﻣﻦ‬ ‫اﻟﻮاﺻﻞ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺑﻴﻦ‬
‫إﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬) ‫ﺗﺪﻋﻰ‬ C ‫أﻓﻖ‬ ‫وﺑﻴﻦ‬ A ‫اﻟﻨﻘﻄﺔ‬
ً‫ﻼ‬‫ﻣﺜ‬ ( C ‫اﻟﻰ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ Angle of Depression A
. (4 - 7 ) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ A C D ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
(4 - 7) ‫اﻟﺸﻜﻞ‬
� � 8 ‫ﻣﺜﺎل‬
‫ﻫـﻲ‬ (‫اﻻﻓﻖ‬ ‫اﻷرض)ﻣﻊ‬ ‫ﻣﻊ‬ ‫اﻟﺨﻴﻂ‬ ‫ﻳﺼﻨﻌﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻛﺎﻧﺖ‬ ‫ﻓﺈذا‬ 30m ‫ﺧﻴﻄﻬﺎ‬ ‫ﻃﻮل‬ ‫ورﻗﻴﺔ‬ ‫ﻃﺎﺋﺮة‬
. ‫اﻻرض‬ ‫ﻋﻦ‬ ‫اﻟﻮرﻗﻴﺔ‬ ‫اﻟﻄﺎﺋﺮة‬ ‫إرﺗﻔﺎع‬ ‫ﺟﺪ‬ 45˚
‫اﻟﻄﻮل‬ ‫وﺣﺪات‬ ‫ﻣﻦ‬ L = ‫اﻻرﺗﻔﺎع‬ ‫أن‬ ‫ﻧﻔﺮض‬
. B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟـــ‬ ‫ﻗـــﺎﺋﻢ‬ A B C ‫اﻟﻤﺜﻠﺚ‬
∴
(4 - 8) ‫اﻟﺸﻜﻞ‬ ∴
A
D
A ‫ﻣﻦ‬ ‫ﻣﻘﺎﺳﻪ‬C ‫ارﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬
C
B ∀
1
2
OPP.
Hyp.
L
30sin 45˚ = ⇒ =
30
2
L= = 21.21m
∀
30m
A
C B
45˚
45˚
L
A ‫اﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬
C ‫ﻣﻦ‬ ‫ﻣﻘﺎﺳﻪ‬

74
C
� � 9 ‫ﻣﺜﺎل‬
60 ˚ ‫ﺗﺴﺎوي‬ ‫ﻗﺎﻋﺪﺗﻬﺎ‬ ‫ﻋﻦ‬ 8m ‫ﺗﺒﻌﺪ‬ ‫اﻷرض‬ ‫ﻋﻠﻰ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﻣﺌﺬﻧﺔ‬ ‫ﻗﻤﺔ‬ ‫إرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫أن‬ ‫اﺻﺪ‬‫ر‬ ‫وﺟﺪ‬
‫؟‬ ‫اﻟﻤﺌﺬﻧﺔ‬ ‫إرﺗﻔﺎع‬ ‫ﻓﻤﺎ‬
: B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ Δ
tan 60˚= ‫ــــــــــ‬
3 = ‫ــــــ‬
.‫اﻟﻤﺌﺬﻧﺔ‬ ‫إرﺗﻔﺎع‬ ‫ﻣﺘﺮ‬ = 8 3 ∴
(4 - 9) ‫اﻟﺸﻜﻞ‬
� � 10 ‫ﻣﺜﺎل‬
‫˚ﻓﻤﺎ‬ 70 ‫اﻻرض‬ ‫ﻋﻠﻰ‬ ‫ﻧﻘﻄﺔ‬ ‫اﻧﺨﻔﺎض‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫أن‬ ‫ﻗﻤﺘﻪ‬ ‫ﻣﻦ‬ ‫اﺻﺪ‬‫ر‬ ‫وﺟﺪ‬ 2350m ‫إرﺗﻔﺎﻋﻪ‬ ‫ﺟﺒﻞ‬
.sin 700
= 0.9396 ‫أن‬ ً‫ﺎ‬‫ﻋﻠﻤ‬ ‫؟‬ ‫اﺻﺪ‬‫ﺮ‬‫واﻟ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﻫﻲ‬
‫اﻻﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻗﻴﺎس‬ = ‫اﻻرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫ﻗﻴﺎس‬
B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ABC
sin 70 ˚= ‫ـــــــ‬
0.9396 = ‫ـــــــــ‬
(4 - 10) ‫اﻟﺸﻜﻞ‬ ∴
AB
8C
A
B˚60
‫ﻤﺌﺬﻧﺔ‬‫اﻟ‬‫رﺗﻔﺎع‬‫ا‬
AB
AC
2350
AC
OPP.
ADJ.
AB
A B C
) ‫اﻟﺸﻜﻞ‬
A
B˚70
C
˚70
2350m
Δ
2350
0.9396
AC = ≅ 2500m
C

75
� � 11 ‫ﻣﺜﺎل‬
‫إﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬‫و‬ 60 ˚ ‫أﻣﺎﻣﻪ‬ ‫ﻋﻤﺎرة‬ ‫أﻋﻠﻰ‬ ‫إرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫أن‬ ‫اﺻﺪ‬‫ر‬ ‫وﺟﺪ‬ ‫ﻣﺘﺮ‬ 7 ‫إرﺗﻔﺎﻋﻪ‬ ‫ﻣﻨﺰل‬ ‫ﺳﻄﺢ‬ ‫ﻣﻦ‬
.‫اﻟﻌﻤﺎرة‬ ‫وإرﺗﻔﺎع‬ ‫واﻟﻌﻤﺎرة‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﺟﺪ‬ ،30˚ ‫ﻗﺎﻋﺪﺗﻬﺎ‬
DAC = AC B
[ ‫اﻻرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ = ‫اﻻﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬]
: B ‫ﻓﻲ‬ ‫اﻟﻘﺎﺋﻢ‬ ABC ‫ﻓﻲ‬
tan 30 ˚ = ‫ــــــ‬
(4 - 11) ‫اﻟﺸﻜﻞ‬ .‫واﻟﻌﻤﺎرة‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬
: D ‫ﻓﻲ‬ ‫اﻟﻘﺎﺋﻢ‬ E A D Δ ‫ﻓﻲ‬
tan 60 ˚ = ‫ــــــ‬
3 = ‫ــــــــــ‬ ⇒ X = 21 m
X+ 7 = ‫اﻟﻌﻤﺎرة‬ ‫إرﺗﻔﺎع‬ ∴
28m =21+7=
∀
7
Y
1
3
7
Y
X
7 3
X
Y
A
C
E
X
D
7
Y˚60
˚30
B ˚30Y
7m
∀
Δ
‫ــــــ‬ = ‫ــــــــ‬ ⇒ Y = 7 3

76
D
x
y
� � 12 ‫ﻣﺜﺎل‬
‫اﻟﻤﻨﻄﺎد‬ ‫ﻧﺤﻮ‬ ‫أﻓﻘﻲ‬ ‫ﻣﺴﺘﻮى‬ ‫ﻓﻲ‬ ‫اﺻﺪ‬‫ﺮ‬‫اﻟ‬ ‫ﺳﺎر‬ ‫وﻟﻤﺎ‬ 30 ˚ ‫ﻫﻲ‬ ‫ﻣﺜﺒﺖ‬ ‫ﻣﻨﻄﺎد‬ ‫إرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫أن‬ ‫اﺻﺪ‬‫ر‬ ‫ﺷﺎﻫﺪ‬
.‫ﻣﺘﺮ‬ ‫أﻗﺮب‬ ‫اﻟﻰ‬ ‫اﻟﻤﻨﻄﺎد‬ ‫إرﺗﻔﺎع‬ ‫ﺟﺪ‬ 45 ˚ ‫ﻫﻲ‬ ‫اﻻرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫أن‬ ‫ﺷﺎﻫﺪ‬ ‫ﻣﺘﺮ‬ 1000 ‫ﻣﺴﺎﻓﺔ‬
: B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ABC
tan45˚= ‫ـــــ‬
1 = ‫ـــــ‬
∴ x = y . . . . . 1
tan 30 ˚= ‫ــــــــــــــــ‬ . . . . .
⇒ 3 y = y + 1000
1.7 y - y = 1000
y = ‫ـــــــ‬
. ‫اﻟﻤﻨﻄﺎد‬ ‫ارﺗﻔﺎع‬ ً‫ا‬‫ﺮ‬‫ﻣﺘ‬ x = 1429 ∴
x
y + 1000
1
3
D
tan 30 ˚=
y
y + 1000
1000
0.7
A
B
C
X
Y˚30
‫0001ﻣﺘﺮ‬
˚45
˚45
Δ
x
y
2
( 4 - 12) ‫اﻟﺸﻜﻞ‬ ‫ـــــــــ‬ = ‫ـــــــــــــــــ‬
= 1428.6

77
: Circular Sector ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ [4-7-2]
(4 - 7) ‫ﺗﻌﺮﻳﻒ‬
‫اﻟﻤﺎرﻳﻦ‬ ‫اﻟﻘﻄﺮﻳﻦ‬ ‫وﺑﻨﺼﻔﻲ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬ ‫ﺑﻘﻮس‬ ‫ﻣﺤﺪد‬ ‫داﺋﺮة‬ ‫ﺳﻄﺢ‬ ‫ﻣﻦ‬ ‫ﺟﺰء‬ ‫ﻫﻮ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬
.‫اﻟﻘﻮس‬ ‫ﺑﻨﻬﺎﻳﺘﻲ‬
‫وﻗﻴﺎﺳﻬﺎ‬ ‫اﻷﺻﻐﺮ‬ ‫اﻟﻘﻄﺎع‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﺑ‬ Central Angle ‫اﻟﻤﺮﻛﺰﻳﺔ‬ AOB ‫ﺗﺴﻤﻰ‬ (4-13) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬
.180 ˚ ‫ﻣﻦ‬ ‫اﻗﻞ‬
× AB ‫اﻟﻘﻮس‬ ‫ﻃﻮل‬ ‫ـــــ‬ = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬
1 . . . . . ‫ـــــ‬ = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬
‫ﻟﻠﻘﻄﺎع‬ ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﻴﺎس‬ ‫أن‬ ‫ﻓﺮﺿﻨﺎ‬ ‫وإذا‬
Q = ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬
L = Q r ⇐ ‫ـــــــ‬ = Q ‫ﻓﺎن‬
: 1 ‫ﻓﻲ‬ ‫وﺑﺎﻟﺘﻌﻮﻳﺾ‬
( 4 - 13 ) ‫اﻟﺸﻜﻞ‬ 2 . . . . . ‫ــــــ‬ = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬
π = ‫اوﻳﺘﻪ‬‫ز‬ ً‫ﺎ‬‫داﺋﺮﻳ‬ ً‫ﺎ‬‫ﻗﻄﺎﻋ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺳﻄﺢ‬ ‫ﻓﺮﺿﻨﺎ‬ ‫إذا‬ : 1 ‫ﻧﺘﻴﺠﺔ‬
‫اﻟﺪاﺋﺮة‬ ‫ﻣﺴﺎﺣﺔ‬ ∴
‫ـــــــ‬ = ‫ــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــــــــــ‬ : 2 ‫ﻧﺘﻴﺠﺔ‬
.‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫ﻟﻠﻘﻄﺎع‬ ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﻴﺎس‬ D ˚‫ﺣﻴﺚ‬ ‫ـــــــ‬ = ‫ـــــــ‬ ∴
‫ـــــــ‬ = ‫ـــــــ‬ = ‫∴ــــــــــــــــــــــــــــــ‬
‫داﺋﺮﺗﻪ‬ ‫ﺳﻄﺢ‬ ‫ﻣﺴﺎﺣﺔ‬ × ‫ــــــــــــــــــــــــــــ‬ = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ∴
∀
1
2
1
2
L
r
1
2
AB
C
O
rr
Q
L
‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬
‫داﺋﺮﺗﻪ‬ ‫ﺳﻄﺢ‬ ‫ﻣﺴﺎﺣﺔ‬
1‫ــــــ‬
2 Qr2
π r2
Q
2 π
Q
2 π
D˚
360˚
‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬
‫داﺋﺮﺗﻪ‬ ‫ﺳﻄﺢ‬ ‫ﻣﺴﺎﺣﺔ‬
Q
2 π
D˚
360˚
‫ﺑﺎﻟﺴﺘﻴﻨﻲ‬ ‫اوﻳﺘﻪ‬‫ز‬ ‫ﻗﻴﺎس‬
360 ˚
‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺤﻴﻂ‬
= r + r + L
= 2 r + L
‫ﻗــﻮس‬ ‫ﻃــــﻮل‬ L ‫ﺣﻴـــﺚ‬
‫ﻃﻮل‬ r ، ‫اﻟــﺪاﺋـــﺮي‬ ‫اﻟﻘﻄﺎع‬
. ‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬
) L r
Q r 2
2
r
1
2 ( 2 ) ×π r2
= π r 2

78
� � 13 ‫ﻣﺜﺎل‬
.8cm ‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 60 ˚ ‫ﻳﺴﺎوي‬ ‫اوﻳﺘﻪ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
Q r2
= ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ∵
× ‫ــــــــ‬ ×64 =
× ‫ــــــــ‬ × 64 =
‫داﺋﺮﺗﻪ‬ ‫ﻣﺴﺎﺣﺔ‬ × ‫ـــــــ‬ = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ : ‫آﺧﺮ‬ ‫ﺣﻞ‬
π × 64 × ‫ـــــــ‬ =
64 × 3 .14 × ‫ـــــــ‬ =
� � 14 ‫ﻣﺜﺎل‬
‫ﻗﻴﺎس‬ ، ‫ﻣﺤﻴﻄﻪ‬ ،‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬ 6cm ‫ﻗﻮﺳﻪ‬ ‫وﻃﻮل‬ 15cm2
‫ﻣﺴﺎﺣﺘﻪ‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬
. ‫ﺑﺎﻟﺴﺘﻴﻨﻲ‬ ‫اوﻳﺘﻪ‬‫ز‬
L r = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ - 1
r + L = ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻄﺎع‬ ‫ﻣﺤﻴﻂ‬ - 2
= 2 × 5 + 6 = 16
‫ﻗﻄﺮﻳﺔ‬ ‫ﻧﺼﻒ‬ ‫اوﻳﺔ‬‫ز‬ 1.2 = ‫ـــــ‬ = Q ⇐ ‫ـــــ‬ = | Q | ∵- 3
‫ـــــــ‬ = ‫ـــــ‬ ⇐ ‫ــــــ‬ = ‫ـــــ‬ ‫ﺛﻢ‬
68 . 7898 ˚ = ‫ـــــــــــــــ‬ = D˚ ∴
1
2
1
2
π 60
180
1
2
3.14
3
D˚
360˚
60˚
360˚
1
6
1
2
1
2
L
r
6
5
Q
D˚
π
180˚
1.2
D˚
3.14
180˚
180˚ × 1.2
3.14
33.49 cm2
15 = × 6 ×r ⇒ r =5
cm
2
33 . 49 cm2

79
: Circular Segment ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ [4-7-3]
(4 - 8) ‫ﺗﻌﺮﻳﻒ‬
.‫اﻟﻘﻮس‬ ‫ذﻟﻚ‬ ‫ﺑﻨﻬﺎﻳﺘﻲ‬ ‫ﻣﺎر‬ ‫ووﺗﺮ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺑﻘﻮس‬ ‫ﻣﺤﺪد‬ ‫داﺋﺮة‬ ‫ﺳﻄﺢ‬ ‫ﻣﻦ‬ ‫ﺟﺰء‬ ‫ﻫﻲ‬ ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬
(4 - 14) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﻤﺮﻛﺰﻳﺔ‬ AOB ‫ﺗﺴﻤﻰ‬
180 ˚ ‫ﻣﻦ‬ ‫اﺻﻐﺮ‬ ‫وﻗﻴﺎﺳﻬﺎ‬ ‫اﻟﺼﻐﺮى‬ ‫اﻟﻘﻄﻌﺔ‬ ‫اوﻳﺔ‬‫ز‬
: ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﻻﻳﺠﺎد‬
(4- 14) ‫اﻟﺸﻜﻞ‬
‫اﻟﺼﻐﺮى‬ ‫اﻟﻘﻄﻌﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻴﺎس‬ Q ‫أن‬ ‫ﻧﻔﺮض‬
OAB Δ ‫ﻣﺴﺎﺣﺔ‬ - ( OACB) ‫اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ = ACB ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ∴
Q r2
= (OACB ‫اﻟﺪاﺋﺮي‬ ‫)اﻟﻘﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ∵
× OA × OB × sin Q = OAB Δ ‫ﻣﺴﺎﺣﺔ‬
× r × r sin Q = OAB ∴
‫ــــــــ‬ - r2
sin Q = AC ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ∴
‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬
. ‫داﺋﺮﺗﻬﺎ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ r ، ‫اﻟﺪاﺋﺮي‬ ‫ﺑﺎﻟﺘﻘﺪﻳﺮ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ Q ‫ﺣﻴﺚ‬
1
2
)
∀
)
1
2
1
2
1
2
1
2
1
2
‫اﻟﺼﻐﺮى‬ ‫اﻟﻘﻄﻌﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻴﺎس‬
AB
C
O
rr
Q
‫ـــــ‬ r2
(Q - sin Q) = ACB
Δ
Q r2

80
� � 15 ‫ﻣﺜﺎل‬
.30˚ ‫اوﻳﺘﻬﺎ‬‫ز‬ ‫وﻗﻴﺎس‬ 12cm ‫داﺋﺮﺗﻬﺎ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬ ‫داﺋﺮﻳﺔ‬ ‫ﻗﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
Q = 0.5236 ⇐ ‫ـــــ‬ = ‫ــــــ‬ ⇐ ‫ـــــ‬ = ‫ــــــ‬
= ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ∵
× 144 × ( 0.5236 -0.5 ) =
1.7cm 2
= × 144 × (0.0236) = ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ∴
� � 16 ‫ﻣﺜﺎل‬
‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬cm2
‫ﻷﻗﺮب‬ ‫ﺟﺪ‬ ،6cm ‫ﻃﻮﻟﻪ‬ ‫وﺗﺮ‬ ‫ﻓﻴﻬﺎ‬ ‫رﺳﻢ‬ ،6cm ‫ﻗﻄﺮﻫﺎ‬ ‫ﻧﺼﻒ‬ ‫داﺋﺮة‬ ‫ﻣﺮﻛﺰ‬ O
. ‫اﻟﺼﻐﺮى‬ ‫اﻟﺪاﺋﺮﻳﺔ‬
(4- 15) ‫اﻟﺸﻜﻞ‬ ‫اﻻﺿﻼع‬ ‫ﻣﺘﺴﺎوي‬ AOB
∴
1.047 = ‫ـــــــ‬ = ‫ـــــــ‬ = Q ⇐ ‫ـــــ‬ = ‫ـــــــ‬ ⇐ ‫ـــــ‬ = ‫ـــــــ‬
= ‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬
× =
18 ( 1.047 -0.865) =
3.276cm2
= 18 (0.182) =
π
180
1
2
Q
D˚
Q
30˚
1
2
1
2
π
180˚
π
180˚
Q
D˚
π
180˚
Q
60˚
π
3
22
21
1
2
1.047 = ‫ـــــــ‬ = ‫ـــــــ‬ =
B
O
6cm
60˚
6cm
‫ـــــ‬ r 2
( Q - sin 30°)
Δ
∀
m AOB = 60 ˚
1
2
‫ـــــ‬ r2
(Q - sinQ)
36
A
(1.047-sin60 ˚)

81
( 4 - 2 ) ‫تمرينات‬
/ 1‫س‬
‫ﻓﻜﺎﻧﺖ‬ ،‫واﺣﺪة‬ ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺒﺮج‬ ‫ﻗﺎﻋﺪة‬ ‫ﻣﻊ‬ ‫ﺗﻘﻌﺎن‬ ‫ﺷﺠﺮﺗﻴﻦ‬ ‫وأﺑﺼﺮ‬ ‫ﺑﺮج‬ ‫أﻋﻠﻰ‬ ‫ﻓﻲ‬ ‫ﺷﺨﺺ‬ ‫وﻗﻒ‬
‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﺟﺪ‬ 50˚ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺸﺠﺮة‬ ‫ﻗﺎﻋﺪة‬ ‫إﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬‫و‬ 70˚ ‫اﻷوﻟﻰ‬ ‫اﻟﺸﺠﺮة‬ ‫ﻗﺎﻋﺪة‬ ‫إﻧﺨﻔﺎض‬ ‫اوﻳﺔ‬‫ز‬
tan 700
= 2. 8 ، tan 500
= 1.2 ‫أن‬ ً‫ﺎ‬‫ﻋﻠﻤ‬ ،30m ‫اﻟﺒﺮج‬ ‫إرﺗﻔﺎع‬ ‫أن‬ ‫اﻟﻌﻠﻢ‬ ‫ﻣﻊ‬ ‫اﻟﺸﺠﺮﺗﻴﻦ‬ ‫ﺑﻴﻦ‬
14.28m/‫ج‬
/ 2‫س‬
‫اﻟﺒﺮج؟‬ ‫إرﺗﻔﺎع‬ ‫ﻓﻤﺎ‬ 300
‫ﻗﻤﺘﻬﺎ‬ ‫إرﺗﻔﺎع‬ ‫اوﻳﺔ‬‫ز‬ ‫أن‬ ‫وﺟﺪ‬ (50m) ‫ﺑﺮج‬ ‫ﻗﺎﻋﺪة‬ ‫ﻋﻦ‬ ‫ﺗﺒﻌﺪ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻦ‬
28.9m /‫ج‬
/ 3‫س‬
.3.2cm ‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 8cm ‫ﻗﻮﺳﺔ‬ ‫ﻃﻮل‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
12.8cm2
/‫ج‬
/ 4‫س‬
.10cm ‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 100° ‫ﺗﺴﺎوي‬ ‫اوﻳﺘﻪ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
87.3cm2
/‫ج‬
/ 5‫س‬
.‫ﻗﻮﺳﻪ‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬ 6cm ‫داﺋﺮﺗﻪ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 37.68cm2
‫ﻣﺴﺎﺣﺘﻪ‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬
12.56cm /‫ج‬
/ 6‫س‬
.450
‫اوﻳﺘﻪ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫ﻓﻴﻬﺎ‬ ‫داﺋﺮي‬ ‫ﻗﻄﺎع‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬ .10cm ‫ﻫﻮ‬ ‫داﺋﺮة‬ ‫ﻣﺤﻴﻂ‬ ‫ﻧﺼﻒ‬
3.98cm2
/‫ج‬
/ 7‫س‬
.8cm‫داﺋﺮﺗﻬﺎ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫وﻃﻮل‬ 60° ‫اوﻳﺘﻬﺎ‬‫ز‬ ‫ﻗﻴﺎس‬ ‫داﺋﺮﻳﺔ‬ ‫ﻗﻄﻌﺔ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
5.81cm2
/‫ج‬

82
�‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﺘﻄﺒﻴﻘﺎت‬ ‫ﻗﻴﻢ‬ ‫إﻳﺠﺎد‬ ‫ﻓﻲ‬ ‫اﻟﺤﺎﺳﺒﺔ‬ ‫إﺳﺘﺨﺪام‬ [4�8�
‫واﻟﺤﺎﺳﺒﺔ‬ ‫اﻟﺪاﺋﺮي‬ ‫واﻟﻘﻴﺎس‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻘﻴﺎس‬ :‫ﻫﻤﺎ‬ ‫ﻟﻠﻘﻴﺎس‬ ‫ﻧﻈﺎﻣﻴﻦ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ ‫ان‬ [4-2] ‫اﻟﺒﻨﺪ‬ ‫ﻓﻲ‬ ‫ﻋﻠﻤﺖ‬
DEG ‫ﻟﻪ‬ ‫ﻳﺮﻣﺰ‬ ‫اﻟﺴﺘﻴﻨﻲ‬ ‫ﻓﺎﻟﻘﻴﺎس‬ ‫اﻟﻴﺪوﻳﺔ‬ ‫اﻟﺤﺎﺳﺒﺔ‬ ‫ﻣﻔﺎﺗﻴﺢ‬ ‫أﻋﻠﻰ‬ ‫ﻳﻼﺣﻆ‬ ‫ﻣﺎ‬ ‫وﻫﻮ‬ ‫اﻟﻨﻈﺎﻣﻴﻦ‬ ‫ﺗﺴﺘﺨﺪم‬
.‫(درﺟﺔ‬DEGREE) ‫ﻟﻜﻠﻤﺔ‬ ً‫ا‬‫ر‬‫أﺧﺘﺼﺎ‬
. ‫ﻗﻄﺮي‬ ‫ﻧﺼﻒ‬ (RADIAN ) ‫ﻟﻜﻠﻤﺔ‬ ً‫ا‬‫ر‬‫أﺧﺘﺼﺎ‬ RAD ‫ﻟﻪ‬ ‫ﻓﻴﺮﻣﺰ‬ ‫اﻟﺪاﺋﺮي‬ ‫اﻟﻘﻴﺎس‬ ‫اﻣﺎ‬
‫اﻷوﻟﻰ‬ ‫ﻓﺎﻟﻀﻐﻄﺔ‬ DRG → ‫اﻟﻤﻔﺘﺎح‬ ‫ﻋﻠﻰ‬ ‫اﻟﻀﻐﻂ‬ ‫ﺑﻌﺪ‬ ‫اﻟﺸﺎﺷﺔ‬ ‫أﻋﻠﻰ‬ ‫ﻓﻲ‬ ‫ان‬‫ﺮ‬‫ﻳﻈﻬ‬ ‫ان‬‫ﺰ‬‫اﻟﺮﻣ‬ ‫وﻫﺬان‬
.‫وﺑﺎﻟﻌﻜﺲ‬ RAD ‫ﺗﻈﻬﺮ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫واﻟﻀﻐﻄﺔ‬ DEG ‫ﺗﻈﻬﺮ‬
.‫اﻟﻈﻞ‬ ‫وﻧﺴﺒﺔ‬ ‫ﺗﻤﺎم‬ ‫اﻟﺠﻴﺐ‬ ‫ﻧﺴﺒﺔ‬ ،‫اﻟﺠﻴﺐ‬ ‫ﻧﺴﺒﺔ‬ ‫ﻋﻠﻰ‬ ‫وﺳﻨﻘﺘﺼﺮ‬ ً‫ﺎ‬‫أﻳﻀ‬ ‫ﻣﻔﺎﺗﻴﺢ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫وﻟﻠﻨﺴﺐ‬
.(sine) ‫اﻟﺠﻴﺐ‬ ‫اﻟﻰ‬ ‫ﻳﺮﻣﺰ‬sin ‫ﻓﺎﻟﻤﻔﺘﺎح‬
.(cosine) ‫ﺗﻤﺎم‬ ‫اﻟﺠﻴﺐ‬ ‫اﻟﻰ‬ ‫ﻳﺮﻣﺰ‬ cos ‫واﻟﻤﻔﺘﺎح‬
.(tangent) ‫اﻟﻈﻞ‬ ‫اﻟﻰ‬ ‫ﻳﺮﻣﺰ‬ tan ‫واﻟﻤﻔﺘﺎح‬
‫اﻟﺤﺎﺳﺒﺔ‬ ‫اﺳﺘﺨﺪام‬ ‫ﻃﺮﻳﻘﺔ‬
. (DRG) ‫ﻋﻠﻰ‬ ‫ﺑﺎﻟﻀﻐﻂ‬ (RAD) ‫اﻟﺪاﺋﺮي‬ ‫أو‬ (DEG) ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻧﻈﺎم‬ ‫ﺗﺤﺪد‬
.‫اﻟﻨﻈﺎم‬ ‫ﺣﺴﺐ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﺗﺪﺧﻞ‬
.‫اﻟﻤﻄﻠﻮﺑﺔ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺐ‬ ‫ﻣﻔﺘﺎح‬ ‫ﻋﻠﻰ‬ ‫ﺗﻀﻐﻂ‬
: ‫ذﻟﻚ‬ ‫ﺗﻮﺿﺢ‬ ‫اﻷﺗﻴﺔ‬ ‫اﻷﻣﺜﻠﺔ‬
� � 17 ‫ﻣﺜﺎل‬
‫ﺟﺪ‬
.‫اﻟﺸﺎﺷﺔ‬ ‫أﻋﻠﻰ‬ DEG ‫ﻟﺘﻈﻬﺮ‬ ‫ﻧﻀﻐﻂ‬ : ‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻨﻈﺎم‬ * (1)
.30 ‫اﻛﺘﺐ‬ *
. 0.5 = ‫اﻟﻨﺎﺗﺞ‬ ‫ﻋﻠﻰ‬ ‫ﻓﺘﺤﺼﻞ‬ (sin) ‫ﻋﻠﻰ‬ ‫اﺿﻐﻂ‬ *
.1
.2
.3
(1) sin 30˚ (2) cos 120˚ (3) tan 350 °
sin( ‫ـ‬ Q) = - sin Q ،
cos (- Q) = cos Q ،
tan (- Q) = - tan Q

83
π
DEG ‫ﻟﺘﻈﻬﺮ‬ ‫ﻧﻀﻐﻂ‬ :‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻨﻈﺎم‬ * (2)
.120 ‫اﻛﺘﺐ‬ *
0.5 = ‫اﻟﻨﺎﺗﺞ‬ ‫ﻋﻠﻰ‬ ‫ﻓﺘﺤﺼﻞ‬ (cos) ‫ﻋﻠﻰ‬ ‫اﺿﻐﻂ‬ *
DEG ‫ﻟﺘﻈﻬﺮ‬ ‫ﻧﻀﻐﻂ‬ :‫اﻟﺴﺘﻴﻨﻲ‬ ‫اﻟﻨﻈﺎم‬ * (3)
0 . 1763 ~ ‫اﻟﻨﺎﺗﺞ‬ ‫ﻓﻴﻜﻮن‬ (tan) ‫ﻋﻠﻰ‬ ‫اﺿﻐﻂ‬ ‫ﺛﻢ‬ 350 ‫اﻛﺘﺐ‬ *
.( ) ( ) ‫ﻓﻴﻜﻮن‬
� � 18 ‫ﻣﺜﺎل‬
‫ﻧﺎﺗﺞ‬ ‫ﺟﺪ‬
RAD ‫ﻟﺘﻈﻬﺮ‬ ‫ﻧﻀﻐﻂ‬ : ‫داﺋﺮي‬ ‫اﻟﻨﻈﺎم‬ ✳
‫ﻟﻸﺳﻮد‬ ‫ﻣﻐﺎﻳﺮ‬ ‫ﺑﻠﻮن‬ ‫وﻳﻜﻮن‬ 2ndf ‫اﻟﻠﻮﺣﺔ‬ ‫ﻋﻠﻰ‬ ً‫ة‬‫ﻋﺎد‬ ‫اﻟﻤﻮﺟﻮد‬ ‫اﻟﻤﻔﺘﺎح‬ ‫ﻋﻠﻰ‬ ‫ﻧﻀﻐﻂ‬ ✳
. ( . . . ً‫ﻼ‬‫ﻣﺜ‬ ‫اﺣﻤﺮ‬ ‫او‬ ‫)اﺻﻔﺮ‬
‫ﻧﺎﺗﺞ‬ = ‫اﻟﻨﺴﺒﺔ‬ ⇐ ‫اﻟﺤﺴﺎﺑﻴﺔ‬ ‫اﻟﻌﻤﻠﻴﺎت‬ ⇐ π :‫ﻣﻔﺘﺎح‬ ‫ﻋﻠﻰ‬ ‫ﻧﻀﻐﻂ‬ ✳
(1)
RAD ‫ﻟﺘﻈﻬﺮ‬ ‫اﺿﻐﻂ‬ *
15.70796327 = 5× ‫اﺿﺮب‬ ⇐ 3.141592654 ‫ﺛﻢ‬ 2ndf ‫ﻧﻀﻐﻂ‬ *
- 0.707106781 ‫ﺛﻢ‬ 3.926990817 = 4 ÷
(2)
.(‫اﻟﺴﺎﻟﺒﺔ‬ ‫اﻻﺷﺎرة‬ ‫)ﻧﺤﺬف‬ cos( - Q) = cos Q ‫أن‬ ‫اﻟﻤﻌﻠﻮم‬ ‫ﻣﻦ‬
.RAD ‫ﻟﺘﻈﻬﺮ‬ ‫اﺿﻐﻂ‬ *
9. 424777961 =3 ×‫⇐اﺿﺮب‬ 3.141592654 = ‫ﺛﻢ‬ 2ndf ‫اﺿﻐﻂ‬ *
cos ‫ﺛﻢ‬
π
- 1 =
-
-
tan - 350° ~ - 0.1763
5π
4
(2) ، sin ‫ـــــــ‬ (1)7π
5
tan ‫ــــــ‬ (3) ، π)cos (- 3
5π
4
sin ‫ـــــــ‬
= sin
π)cos (-3
tan ( - Q ) = - tan Q
INV‫أو‬

84
π
(3)
.RAD ‫ﻟﺘﻈﻬﺮ‬ ‫اﺿﻐﻂ‬ *
21 .9114858 = 7 × ‫اﺿﺮب‬ ⇐ 3.141592654 = ‫ﺛﻢ‬ 2ndf ‫*اﺿﻐﻂ‬
. 3.07763537 = ‫اﺿﻐﻂ‬ ‫ﺛﻢ‬ 4 .398229715 = 5 ÷ ⇐
: ‫اﻟﺤﺎﺳﺒﺔ‬ ‫ﺑﺎﺳﺘﺨﺪام‬ ‫ﻳﺄﺗﻲ‬ ‫ﻣﺎ‬ ‫ﺟﺪ‬ ‫ﺗﻤﺮﻳﻦ‬
(-400°) (2) ‫ــــ‬ (1)
،
‫اﻻﺟﺎﺑﺔ‬
0.5 (1)
0.766044443 (2)
- 0.267949192 (3)
- 0.588 (4)
- 0.5 (5)
-3.077683537 (6)
π
6
7π
5
tan ‫ــــــ‬
tan
sincostan (-15 ˚) (3)tan (-36 ˚) (4)
2π
3cos (5)8π
5tan (6)

85
� Solution of Right Angled Triangle ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﺣﻞ‬ [4�9�
‫ﻗﻴﻢ‬ ‫إﻳﺠﺎد‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﺑﺤﻞ‬ ‫وﻳﻘﺼﺪ‬ [‫زواﻳﺎ‬ ‫وﺛﻼث‬ ‫أﺿﻼع‬ ‫]ﺛﻼﺛﺔ‬ ‫ﻋﻨﺎﺻﺮ‬ ‫ﺳﺘﺔ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺜﻠﺚ‬ ‫ﻛﻞ‬ ‫ﻳﺸﺘﻤﻞ‬
.‫اﻟﻤﺠﻬﻮﻟﺔ‬ ‫ﻋﻨﺎﺻﺮه‬
� �19 ‫ﻣﺜﺎل‬
: ‫أوﺟﺪ‬ tan 22 °= 0. 4 ‫ﻛﺎن‬ ‫إذا‬
sin 22° ، cos 22° (1)
cos 68°، sin 68° (2)
tan 22 ° = ‫ــــــــــ‬ = ‫ــــــــ‬ = ‫ــــــــ‬
2k = ‫اﻟﻤﻘﺎﺑﻞ‬ ∴
5k = ‫اﻟﻤﺠﺎور‬ ∴
‫ﻓﻴﺜﺎﻏﻮرس‬‫ﻣﺒﺮﻫﻨﺔ‬......
4K2
+ 25K2
= (Ac)2
AC = 29 K
22° = ‫ــــــــ‬ = ‫ــــــــ‬ = ‫ــــــــــ‬ (1)
22°= ‫ــــــــ‬ = ‫ــــــــ‬ = ‫ــــــــــ‬
68° = (90° -22°) = 22° = (2)
2
5
4
10
BC
AC
AB
AC
( AB )2
+ ( BC )2
=( AC )2
2k
29k
5k
29k
2
29
5
29
sin
cos
cossin
5 KB
C
A
68°
22°
2 K
sin
cos 68° = cos( 90° - 22°) = sin 22°= 2
29
29 k
5
29

86
K 2
� � 20 ‫ﻣﺜﺎل‬
tan C ، sinA, ‫ﺟﺪ‬ .B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ABC ‫ﻓﻲ‬ cos C = ‫ــــــــ‬ ‫أن‬ ‫ﻋﻠﻤﺖ‬ ‫إذا‬
.cosA
: B ‫ﻓﻲ‬ ‫اﻟﻘﺎﺋﻢ‬ ABC ‫ﻧﺮﺳﻢ‬
cos C = ‫ــــــــ‬ = ‫ــــــــــ‬
(‫)ﻓﻴﺜﺎﻏﻮرس‬ ∵
K 2
= (AB)2
+ 25 ∴
∴ (AB)2
= 144 K 2
⇒ AB = 12K
tan C = ‫ــــــــ‬ = ‫ــــــــ‬
cosA = ‫ــــــــ‬ = ‫ــــــــ‬
� � 21 ‫ﻣﺜﺎل‬
: ‫ﺟﺪ‬ AB = 7 cm ، AC = 24 cm ‫ﻓﻴﻪ‬ A ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ ABC
sin C ، sin B ، tan C ، cos B
( BC )2
=( A B)2
+(A C)2
(B C )2
= (7) 2
+(24) 2
= 49 + 576 = 625
BC = 25 cm ∴
5
13
5k
13k
‫اﻟﻮﺗﺮ‬
12k
5k
12
5
12k
13k
12
13
Δ
(AC) 2
= (AB)2
+ (BC) 2
169
C
A
B
12K
13K
5K
5k
13k
5
13sin A= =
C
B
A
25cm
24cm
7cm

87
∴ sin C = ‫ــــــــ‬ ، sin B = ‫ــــــــ‬
tan C= ‫ــــــــ‬ ، cos B = ‫ــــــــ‬
� � 22 ‫ﻣﺜﺎل‬
AC = 6 cm AB = 3 cm ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ . B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ABC ‫اﻟﻤﺜﻠﺚ‬ ‫ﺣﻞ‬
(AC)2
= (AB)2
+ (BC)2
36 = 9 + (BC)2
BC = 3 3
‫اﻟﺒﺎﻗﻴﺔ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺳﻨﺠﺪ‬ ‫واﻻن‬ ، ‫اﻻﺿﻼع‬ ‫اﻃﻮال‬ ‫اﻳﺠﺎد‬ ‫اﺳﺘﻜﻤﻠﻨﺎ‬
tan C = ‫ــــــــ‬ = ‫ ⇒ ــــــــ‬
m < A = 90° - 30° = 60°
7
25
7
24
24
25
7
25
C = 30°3
3 3
1
3
A
C
6 cm
3cm
3 3 cm
60ْ
30ْ B
:‫ﻧﺴﺘﺨﺪم‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫اﻟﻘﺎﺋﻢ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﺣﻞ‬ ‫ﻓﻲ‬
tanQ, cosQ, sinQ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ *
‫ﻓﻴﺜﺎﻏﻮرس‬ ‫ﻣﺒﺮﻫﻨﺔ‬ ‫ﻧﺴﺘﺨﺪم‬ *
‫ﺳﺆال‬ ‫ﻛﻞ‬ ‫ﻃﺒﻴﻌﺔ‬ ‫وﺣﺴﺐ‬

88
( 4 - 3 ) ‫تمرينات‬
/ 1‫س‬
. cos C ، tan C ، sin A : ‫ﺟﺪ‬ sin C = ‫ـــــــ‬ ‫ﻓﻴﻪ‬ B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ ABC
/ 2‫س‬
‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬ AB = 25 cm ، B C = 24 cm ‫ﻓﻴﻪ‬ C ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ AB C
.‫اﻟﻤﻌﻄﺎة‬ ‫اﻟﻤﻌﻠﻮﻣﺎت‬ ‫وﺑﺎﺳﺘﺨﺪام‬ sin2
B + cos2
B
/ 3‫س‬
sin Q ، tan Q ‫ﻓﺄوﺟﺪ‬ cos Q = ‫ــــــــ‬ ‫ﻛﺎن‬ ‫إذا‬
/ 4‫س‬
‫ﻛﺎﻧﺖ‬ ‫ﻓﺎذا‬ ‫ﺷﺎﻗﻮﻟﻲ‬ ‫ﺣﺎﺋﻂ‬ ‫ﻋﻠﻰ‬ ‫اﻵﺧﺮ‬ ‫وﻃﺮﻓﻪ‬ ‫أﻓﻘﻴﺔ‬ ‫أرض‬ ‫ﻋﻠﻰ‬ ‫اﻷﺳﻔﻞ‬ ‫ﻃﺮﻓﻪ‬ ‫ﻣﺮﺗﻜﺰ‬ ‫ﻣﺘﺮ‬ 10 ‫ﻃﻮﻟﻪ‬ ‫ﺳﻠﻢ‬
‫اﻟﺤﺎﺋﻂ؟‬ ‫ﻋﻦ‬ ‫اﻷﺳﻔﻞ‬ ‫وﻃﺮﻓﻪ‬ ‫اﻷرض‬ ‫ﻋﻦ‬ ‫اﻷﻋﻠﻰ‬ ‫ﻃﺮﻓﻪ‬ ‫ﺑﻌﺪ‬ ‫°03ﻓﻤﺎ‬ ‫واﻷرض‬ ‫اﻟﺴﻠﻢ‬ ‫ﺑﻴﻦ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
( 3 =1. 73 ) ‫اﺳﺘﻌﻤﻞ‬
/ 5‫س‬
.‫ﻣﻨﻄﻘﺘﻪ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬ A B = 20cm ، m< C A B = 60 ° ‫ﻓﻴﻪ‬ C ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ A B C
/ 6‫س‬
:‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬
( A ) ‫ــــــ‬ tan2
30 °+2sin 60°+3tan45°+cos2
30°-tan60°
(B) cos2
45°sin60°tan60°cos2
30 ° .
(C) sin 120° ، cos 135°، tan 150°.
/ 7‫س‬
: ‫اﻟﻤﺠﺎور‬ ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬
‫ﻣﻨﺤﺮف‬ ‫ﺷﺒﻪ‬ ABCD
AD=BC ‫ﻓﻴﻪ‬
، ( ‫اﻟﺴﺎﻗﻴﻦ‬ ‫)ﻣﺘﺴﺎوي‬
،DC=20cm، AB=14cm
m<CDA ‫،ﺟﺪ‬AD=6cm
8
17
4
5
3
4
A B
D C

898989
5 Vectors ‫ﺍﻟﻤﺘﺠﻬﺎﺕ‬ : ‫ﺍﻟﺨﺎﻣﺲ‬ ‫ﺍﻟﻔﺼﻞ‬
ً‫ا‬‫وجبري‬ ً‫ا‬‫هندسي‬ ‫المتجه‬ ‫مفهوم‬ [5-1]
‫المقيد‬ ‫المتجه‬ [5-2]
‫واتجاهه‬ ‫المتجه‬ ‫ﻃول‬ ‫إيجاد‬ [5-3]
‫حقيقي‬ ‫بعدد‬ ‫وضربها‬ ‫المتجهات‬ ‫جمع‬ [5-4]
‫اﻟﻤﺴﺘﻮى‬ ‫ﻓﻲ‬ ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫اﻋﻄﺎء‬ [5-5]

a = (x , y) a ‫اﻟﻤﺘﺠﻪ‬
a = x 2
+ y 2
a ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬
0 = (0 , 0 ) ‫اﻟﺼﻔﺮي‬ ‫اﻟﻤﺘﺠﻪ‬ *
u1
= (1 , 0 ) , u2
= ( 0 , 1 ) u1
, u2
‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬
:‫ان‬ ‫على‬ ً‫ا‬‫ر‬‫قاد‬ ‫يكون‬ ‫الفصل‬ ‫هذا‬ ‫نهاية‬ ‫في‬ ‫للطالب‬ ‫ينبﻐي‬
.ً‫ا‬‫هندسي‬ ‫المتجه‬ ‫على‬ ‫يتعرف‬ -
ً‫ا‬‫جبري‬ ‫المتجه‬ ‫على‬ ‫يتعرف‬ -
‫المقيد‬ ‫المتجه‬ ‫على‬ ‫يتعرف‬ -
‫المقيد‬ ‫المتجه‬ ‫ﻃول‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫المقيد‬ ‫المتجه‬ ‫اتجاه‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫المتجهات‬ ‫جمع‬ ‫من‬ ‫يتمكن‬ -
‫حقيقي‬ ‫بعدد‬ ‫المتجه‬ ‫ضرب‬ ‫من‬ ‫يتمكن‬ -
‫الوحدة‬ ‫متجهي‬ ‫على‬ ‫يتعرف‬ -
‫الوحدة‬ ‫متجهي‬ ‫بداللة‬ ‫المتجه‬ ‫وضع‬ ‫من‬ ‫يتمكن‬ -

90
Vectors ‫اﻟﻤﺘﺠﻬﺎت‬ : ‫اﻟﺨﺎﻣﺲ‬ ‫اﻟﻔﺼﻞ‬
(( ‫واﻟﺠﺒﺮي‬ ‫))اﻟﻬﻨﺪﺳﻲ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻣﻔﻬﻮم‬ [5�1�
‫وﻏﻴﺮﻫﺎ‬ ‫واﻟﻤﺴﺎﻓﺔ‬ ‫واﻟﺤﺠﻢ‬ ‫واﻟﺰﻣﻦ‬ ‫واﻟﻜﺘﻠﺔ‬ ‫اﻟﻄﻮل‬ ‫ﻣﺜﻞ‬ ‫واﻟﺮﻳﺎﺿﻴﺔ‬ ‫اﻟﻔﻴﺰﻳﺎﺋﻴﺔ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫ﺑﻌﺾ‬ � ‫ﻣﻘﺪﻣﺔ‬
‫اﻟﻌﺪدﻳﺔ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫ﺗﺴﻤﻰ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫ﻫﺬه‬ ‫ﻣﺜﻞ‬ ،‫ﻓﻘﻂ‬ ‫ﻣﻘﺪارﻫﺎ‬ ‫ﻋﻠﻰ‬‫ﻳﺪل‬ ‫ﻋﺪد‬ ‫ﺑﺬﻛﺮ‬ ً‫ﻼ‬‫ﻛﺎﻣ‬ً‫ا‬‫ﺗﺤﺪد‬ ‫ﺗﺘﺤﺪد‬
‫اﻟﻰ‬ ‫ﺑﺎﻻﺿﺎﻓﺔ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻳﻜﻮن‬ ‫اﺣﺔ‬‫ز‬‫واﻷ‬ ‫واﻟﺴﺮﻋﺔ‬ ‫اﻟﻘﻮة‬ ‫ﻣﺜﻞ‬ ‫أﺧﺮى‬ ‫وﻛﻤﻴﺎت‬ .‫اﻟﻤﺘﺠﻬﺔ‬ ‫ﻏﻴﺮ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫أو‬
‫ﻧﺸﺄت‬ .‫اﻟﻤﺘﺠﻬﺔ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫ﺗﺴﻤﻰ‬ ‫اﻟﻜﻤﻴﺎت‬ ‫ﻫﺬه‬ ‫ﻣﺜﻞ‬ ً‫ﻼ‬‫ﻛﺎﻣ‬ ً‫ا‬‫ﺗﺤﺪﻳﺪ‬ ‫ﺗﺤﺪﻳﺪﻫﺎ‬ ‫ﻓﻲ‬ ً‫ﺎ‬‫ﺿﺮورﻳ‬ ‫اﻟﻤﻘﺪار‬
‫اﻟﻘﻄﻌﺔ‬ ‫وإﺳﺘﺨﺪﻣﺖ‬ ،‫وﻏﻴﺮﻫﺎ‬ ‫اﺣﺔ‬‫ز‬‫واﻹ‬ ‫واﻟﺴﺮﻋﺔ‬ ‫اﻟﻘﻮة‬ ‫ﻟﺘﻤﺜﻴﻞ‬ ‫اﻟﻤﻴﻜﺎﻧﻴﻚ‬ ‫ﻋﻠﻢ‬ ‫ﻓﻲ‬ ً‫ﻼ‬‫أﺻ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻓﻜﺮة‬
‫ﻧﻘﻄﺔ‬ ‫ﺗﺴﻤﻰ‬ B ‫ﻣﺜﻞ‬ ‫أﺧﺮى‬ ‫ﻧﻘﻄﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺒﺪء‬ ‫ﻧﻘﻄﺔ‬ ‫ﺗﺴﻤﻰ‬ A ‫ﻣﺜﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺘﺠﻬﺔ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬
‫ﻣﻮﺟﻬﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫أن‬ ‫اﻟﺴﻬـــﻢ‬ ‫ﻳﻌﻨﻲ‬ ‫ﺣﻴﺚ‬ A B ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻠﻤﺘﺠﻪ‬ ‫ﻋﺎدة‬ ‫وﻳﺮﻣﺰ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻟﺘﻤﺜﻴﻞ‬ ‫اﻻﻧﺘﻬﺎء‬
‫إﺗﺠﺎﻫﺎن‬ ‫ﻫﻨﺎك‬ (‫وﻧﻬﺎﻳﺘﻪ‬ ‫ﺑﺪاﻳﺘﻪ‬ ‫ﻣﻌﺮﻓﺔ‬ ‫)ﻣﻊ‬ a ‫ﻣﺜﻞ‬ ‫واﺣﺪ‬ ‫ﺑﺤﺮف‬ ‫ﻟﻠﻤﺘﺠﻪ‬ ‫ﻳﺮﻣﺰ‬ ‫وﻗﺪ‬ B ‫اﻟﻰ‬ A ‫ﻣﻦ‬
-: ‫اﻟﻤﺘﺠﻬﺎت‬ ‫اﺳﺔ‬‫ر‬‫ﻟﺪ‬
‫ﻫﻨﺪﺳﻲ‬ (1)
‫ﺟﺒــﺮي‬ (2)
‫ﻷﺟﻞ‬ ‫اﻟﻬﻨﺪﺳﻲ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻦ‬ ‫ﻣﺴﺘﻔﻴﺪﻳﻦ‬ ‫اﻟﺠﺒﺮي‬ ‫اﻻﺗﺠﺎه‬ ‫ﻋﻠﻰ‬ ‫اﻟﻔﺼﻞ‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫اﺳﺘﻨﺎ‬‫ر‬‫د‬ ‫ﻓﻲ‬ ‫وﺳﻨﺆﻛﺪ‬
.‫اﻟﺘﻮﺿﻴﺢ‬
A B
a

91
‫أﺳﻠﻔﻨـــﺎ‬ ‫ﻛﻤــــﺎ‬ ‫ﻣﻮﺟﻬﺔ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻗﻄﻌﺔ‬ ‫ﻳﻌﻨﻲ‬ ‫اﻟﻬﻨﺪﺳﻴﺔ‬ ‫اﻟﻨـﺎﺣﻴﺔ‬ ‫ﻣﻦ‬ :‫اﻟﻤﺘﺠﻪ‬ ‫أﺳﺎﺳﻴﺔ‬ ‫ﻣﻔﺎﻫﻴﻢ‬
. ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻣﺘﺠﻬﺎت‬ ‫ﺗﻤﺜﻞ‬ AB ، CD ، EF ‫ﻓﺎﻟﻘﻄﻊ‬
( 5 - 1 ) ‫اﻟﺸﻜﻞ‬
‫اﻻﺗﺠﺎه‬ ‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫ﻟﻠﻤﺘﺠﻬﻴﻦ‬ ‫ﻳﻜﻮن‬ ‫ﻗﺪ‬ ،‫ﻣﺘﻮازﻳﺘﻴﻦ‬ ‫ﻗﻄﻌﺘﺎﻫﻤﺎ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ �‫اﻟﻤﺘﻮازﻳﺎن‬ ‫اﻟﻤﺘﺠﻬﺎن‬
‫ﻧﻔﺲ‬ ‫وﻟﻬﻤﺎ‬ C D ‫ﻳﻮازي‬ A B ‫ان‬ ‫ﻧﻼﺣﻆ‬ (5-2) ‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ . ‫ﻣﺘﻌﺎﻛﺴﻴﻦ‬ ‫ﺑﺎﻻﺗﺠﺎه‬ ‫ﻳﻜﻮﻧﺎن‬ ‫وﻗﺪ‬ ‫ﻧﻔﺴﻪ‬
‫اﻻﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫ﻣﺘﻌﺎﻛﺴﺎن‬ ‫اﻧﻬﻤﺎ‬ ‫ﻛﻤﺎ‬ E F ‫ﻳﻮازي‬ A B ‫وﻟﻜﻦ‬ ‫اﻻﺗﺠﺎه‬
( 5 - 2 ) ‫اﻟﺸﻜﻞ‬
‫ﻧﻔﺴﻪ‬ ‫واﻻﺗﺠﺎه‬ ‫ﻧﻔﺴﻪ‬ ‫اﻟﻄﻮل‬ ‫ﻟﻬﻤﺎ‬ ‫ﻛﺎن‬ ‫إذا‬ �‫اﻟﻤﺘﻜﺎﻓﺌـﺎن‬ ‫اﻟﻤﺘﺠﻬﺎن‬
BF
D
CE
F
E D B
C A
A

92
Conorlical Vector� ‫اﻟﻤﻘﻴﺪ‬ ‫اﻟﻤﺘﺠﻪ‬ [5�2�
‫ﻣﻦ‬ ً‫ﻻ‬‫ﻓﺒﺪ‬ ‫ﻟﺬا‬ ، (0 , 0) ‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﻳﺒﺘﺪئ‬ ‫ﻳﻜﺎﻓﺌﻪ‬ ‫وﺣﻴﺪ‬ ‫ﻣﺘﺠﻪ‬ ‫ﻳﻮﺟﺪ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫ﻣﺘﺠﻪ‬ ‫ﻟﻜﻞ‬
‫اﻟﻤﻜﺎﻓﻲء‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺳﻨﺘﺨﺬ‬ ،‫واﻻﺗﺠﺎه‬ ‫اﻟﻄﻮل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺘﺴﺎوﻳﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ‫ﻣﻨﺘﻪ‬ ‫ﻏﻴﺮ‬ ‫ﻋﺪد‬ ‫ﻣﻦ‬ ‫اﻟﺘﻌﺎﻣﻞ‬
‫ﺑﺎﻟﻤﺘﺠﻪ‬ ‫اﻻﺻﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫ﻳﺒﺘﺪئ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻳﺴﻤﻰ‬ ،ً‫ﺎ‬‫ﺟﻤﻴﻌ‬ ‫ﻋﻨﻬﺎ‬ ً‫ﻼ‬‫ﻣﻤﺜ‬ ‫اﻷﺻﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫ﻳﺒﺘﺪئ‬ ‫واﻟﺬي‬ ‫ﻟﻬﺎ‬
‫أو‬ ‫اﻟﺤﺮ‬ ‫)اﻟﻤﺘﺠﻪ‬ ‫اﻻﺻﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫اﻟﻤﺮﺗﺒﻄﺔ‬ ‫ﻏﻴﺮ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﺑﻘﻴﺔ‬ ‫وﺗﺴﻤﻰ‬ .‫اﻟﻤﻘﻴﺪ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫أو‬ ‫اﻟﻘﻴﺎﺳﻲ‬
.(‫اﻟﻄﻠﻴﻖ‬ ‫اﻟﻤﺘﺠﻪ‬
: ‫أن‬ ‫ﻻﺣﻆ‬
‫ﻣﻘﻴﺪان‬ ‫ﻣﺘﺠﻬﺎن‬ O F ، O E
‫ﻃﻠﻴﻘﺎن‬ ‫ﻣﺘﺠﻬﺎن‬ C D ، A B ‫ﺑﻴﻨﻤﺎ‬
( 5 - 3 ) ‫اﻟﺸﻜﻞ‬
: ‫وﺗﻤﺜﻴﻠﻬﺎ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ [5-2-1]
‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬ ‫زوج‬ ‫وﻛﻞ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫اﻟﻤﺘﻌﺎﻣﺪ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫ﺑﻨﻘﻄﺔ‬ (3 , 4) ‫اﻟﺰوج‬ ‫ﻣﺜﻠﻨﺎ‬ ‫ﻟﻘﺪ‬
‫ﻋﻠﻰ‬ C ، B ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻳﺘﻤﺜﻼن‬ (3 , 2) ،(5 , 3) ‫اﻟﻤﺮﺗﺒﺎن‬ ‫ﻓﺎﻟﺰوﺟﺎن‬ ‫واﺣﺪة‬ ‫ﺑﻨﻘﻄﺔ‬ ‫ﺗﻤﺜﻴﻠﻪ‬ ‫ﻧﺴﺘﻄﻴﻊ‬
. ‫اﻟﺘﻮاﻟﻲ‬
‫اﻟﻤﺮﺗﺐ‬ ‫اﻟﺰوج‬ ‫ﺗﻤﺜﻴﻞ‬ ‫وﻧﺴﺘﻄﻴﻊ‬
‫ﺑﻘﻄﻌﺔ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬
‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﺪاﻳﺘﻬﺎ‬ ‫ﻣﺘﺠﻬﻪ‬
‫اﻟﻤﻌﻠﻮم‬ ‫اﻟﻤﺮﺗﺐ‬ ‫اﻟﺰوج‬ ‫وﻧﻬﺎﻳﺘﻬﺎ‬
OC ، OB ، OA ‫اﻟﻤﻮﺟﻬﺔ‬ ‫ﻓﺎﻟﻘﻄﻊ‬
( 5 - 4 ) ‫اﻟﺸﻜﻞ‬
Y
D
C
B
A F
X
E
O
Y
X
A(3,4)
B(5,3)C(3,2)
O

93
‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬ ‫ﺑﺰوج‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺳﻨﻤﺜﻞ‬ ‫اﻷﺳﺎس‬ ‫ﻫﺬا‬ ‫ﻋﻠﻰ‬ . (3,2) ، (5,3) ، (3,4) ‫اﻟﻤﺮﺗﺒﺔ‬ ‫اﻷزواج‬ ‫ﺗﻤﺜﻞ‬
: ‫ﻧﻜﺘﺐ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬
‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻋﻠﻰ‬ ‫اﺳﺘﻨﺎ‬‫ر‬‫د‬ ‫ﻓﻲ‬ ‫ﻧﻘﺘﺼﺮ‬ ‫ﺳﻮف‬ ‫ﻷﻧﻨﺎ‬
‫اﻷﺻــﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫ﺗﺒﺘﺪئ‬ ‫ﻛﻠﻬﺎ‬ ‫ﻟﺬا‬ ،‫ﻓﻘﻂ‬ ‫اﻟﻤﻘﻴﺪة‬
. ‫ﻓﻘﻂ‬ ‫اﻟﻨﻬﺎﺋﻴﺔ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫ﻓﻨﺬﻛﺮ‬
( 5 - 5 ) ‫اﻟﺸﻜﻞ‬
� ‫واﺗﺠﺎﻫﻪ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬ [5�3�
: ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬ [5-3-1]
.‫أﻧﺘﻬﺎﺋﻪ‬ ‫وﻧﻘﻄﺔ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺑﺪاﻳﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﻫﻲ‬
. || AB || ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ AB ‫ﻃﻮل‬ ‫ﻳﺴﺎوي‬ AB ‫ﻓﻄﻮل‬
( 5-6 ) ‫اﻟﺸﻜﻞ‬
(5-1) ‫ﺗﻌﺮﻳﻒ‬
: ‫ﻓﺎن‬ ‫ﺣﻴﺚ‬ ‫ﻣﺘﺠﻬﺎ‬ A ‫ﻛﺎن‬ ‫اذا‬
OA = || A || =O A = x2
+ y2
(5 - 6) ‫اﻟﺸﻜﻞ‬ ‫ﻻﺣﻆ‬
OA = A = (x , y)
A = (x,y)
Y
X
A(x,y)
L
O(0,0)
Y
X
A(x,y)
Y
X
O(0,0)
A

94
� � 1 ‫ﻣﺜﺎل‬
: ‫اﻷﺗﻴﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬
( -12 , -9 ) ، ( ‫ـــــــــــ‬ , ‫ـــــــــــ‬ ) ، (3,4)
: ‫ﻫﻮ‬ (3,4) ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬
: ‫ﻫﻮ‬ ( ‫ـــــــــ‬ , ‫ـــــــــ‬ ) ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬
: ‫ﻫﻮ‬ ( -12 , -9 ) ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬
(5-2) ‫ﺗﻌﺮﻳﻒ‬
‫وﻧﻬﺎﻳﺘﻪ‬ ‫ﺑﺪاﻳﺘﻪ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻻن‬ ‫اﻟﺼﻔﺮي‬ ‫ﺑﺎﻟﻤﺘﺠﻪ‬ (0,0) ‫اﻟﻤﺘﺠﻪ‬ ‫ﻳﺴﻤﻰ‬ : Zero Vector ‫اﻟﺼﻔﺮي‬ ‫اﻟﻤﺘﺠﻪ‬
. ‫ﺻﻔﺮ‬ = || 0 || = 0 ‫وﻃﻮل‬ ، 0 ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ . ‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻫﻲ‬
(5-3)‫ﺗﻌﺮﻳﻒ‬
‫ﻛﺎن‬ ‫إذا‬ ‫وﻓﻘﻂ‬ ‫أذا‬ ‫ﻣﺘﺴﺎوﻳﺎن‬ ‫أﻧﻬﻤﺎ‬ (x1
, y1
) ‫و‬ (x2
,y2
) ‫ﻟﻠﻤﺘﺠﻬﻴﻦ‬ ‫ﻳﻘﺎل‬ :‫اﻟﻤﺘﺴﺎوﻳﺎن‬ ‫اﻟﻤﺘﺠﻬﺎن‬
.x1
= x2
، y1
= y2
(5-4) ‫ﺗﻌﺮﻳﻒ‬
.‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻊ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻳﺼﻨﻌﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ :‫اﻟﻤﺘﺠﻪ‬ ‫اﺗﺠﺎه‬
7 2
10
2
10
( 3 )2
+ ( 4 )2
= 9 + 16 = 5
(‫)ـــــــ‬2
+ (‫)ـــــــ‬2
= ‫ــــــ‬ + ‫ــــــ‬ = ‫ـــــــ‬ = 12
10
7 2
10
2
100
98
100
100
100
7 2
10
2
10
(-12)2
+ (-9)2
= 144 + 81 = 225 = 15

95
: ‫اﻟﻤﺘﺠﻪ‬ ‫اﺗﺠﺎه‬ ‫أﻳﺠﺎد‬ [5-3-2]
‫ﻣﻘﺎﺳﺔ‬ ‫ﺣﻴﺚ‬ Q ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﺑﻘﻴﺎس‬ ‫ﻳﻌﺮف‬ A ‫إﺗﺠﺎه‬ ‫ﻓﺄن‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ ‫ﻛﺎن‬ ‫إذا‬
A ‫اﻟﻤﺘﺠﻪ‬ ‫اﻟﻰ‬ ‫ﺟﺐ‬ ‫اﻟﻤﻮ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ‫اﻟﺴﺎﻋﺔ‬ ‫ﻋﻘﺎرب‬ ‫ﺣﺮﻛﺔ‬ ‫ﻻﺗﺠﺎه‬ ‫ﻣﻌﺎﻛﺲ‬ ‫ﺑﺄﺗﺠﺎه‬
.‫اﺗﺠﺎﻫﻪ‬ ‫ﺗﻌﺮﻳﻒ‬ ‫ﻻﻳﻤﻜﻦ‬ ‫اﻟﺼﻔﺮي‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ان‬ ‫ﻻﺣﻆ‬
cos Q= ‫ــــــــــــــ‬ , sin Q = ‫ـــــــــــــــ‬
� � 2 ‫ﻣﺜﺎل‬
‫واﺗﺠﺎه‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬
|| OB || = ( 3 )2
+ ( -1 )2
= 3 + 1 = 2
‫ﻳﺤﺪدﻫﺎ‬ ‫اﻟﺘﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﻴﺎس‬ ‫ﻳﺴﺎوي‬ Q ‫أن‬ ‫ﻧﻔﺮض‬
‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻊ‬ OB ‫اﻟﻤﺘﺠﻪ‬
‫ﻓﻴﻜﻮن‬
Q ‫ان‬ ‫ﻧﻼﺣﻆ‬ (5-7) ‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬
‫اﺑﻊ‬‫ﺮ‬‫اﻟ‬ ‫اﻟﺮﺑﻊ‬ ‫ﻓﻲ‬ ‫ﺗﻘﻊ‬
2 - ‫ـــــــ‬ = ‫ـــــــ‬ : ‫ﻫﻮ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫اﺗﺠﺎه‬
( 5 - 7 ) ‫اﻟﺸﻜﻞ‬
A = (x , y)0 ≤ Q < 2 π
x
x2
+ y2
y
x2
+ y2
OB = ( 3 , -1)
cos Q = ‫ــــــــ‬
sin Q = ‫ــــــــ‬
3
2
-1
2
π π π11
66
Y
X
B
-1
Q
O
3
( 3 , -1)

96
� � 3 ‫ﻣﺜﺎل‬
( ‫ـــــــ‬ , ‫ـــــــ‬ ) ‫اﻟﻤﺘﺠﻪ‬ ‫إﺗﺠﺎه‬ ‫ﺟﺪ‬
( ‫ـــــــ‬ , ‫ـــــــ‬ ) ‫اﻟﻤﺘﺠﻪ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫ﺗﺴﺎوي‬ Q ‫أن‬ ‫ﻧﻔﺮض‬
(5 - 8) ‫اﻟﺸﻜﻞ‬
cos Q = ‫ـــــــــــــــــــــــــــ‬ = ‫ــــــــ‬
sin Q = ‫ــــــــــــــــــــــــــــــ‬ = ‫ـــــــــ‬
:‫)8-5(ﻧﻼﺣﻆ‬ ‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬
‫ـــــــ‬ ‫ﻓﺘﻜﻮن‬ ‫اﻻول‬ ‫اﻟﺮﺑﻊ‬ ‫ﻓﻲ‬ ‫ﺗﻘﻊ‬ Q
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
( )2
+ ( )2
1
2
1
2
( )2
+ ( )2
π
4
y
X
1
2
1
2
45ْ
0
11
22
1
2
1
2
( 1
2
1
2
(،
1
2
1
2

97
� � 4 ‫ﻣﺜﺎل‬
‫ـــــ‬ ‫وإﺗﺠﺎﻫﻪ‬ ‫وﺣﺪات‬ 5 = ‫ﻃﻮﻟﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺟﺪ‬
( x , y ) = ‫اﻟﻤﺘﺠﻪ‬ ‫ﻧﻔﺮض‬
cos Q = ‫ـــــــــ‬ ⇒ cos ‫ـــــــ‬ = ‫ـــــــــ ⇒ ـــــــ‬ = ‫ـــــــــ‬
∴ x = ‫ــــــــ‬
sin Q = ‫ـــــــــ‬ ⇒ sin ‫ـــــــ‬ = ‫ـــــــــ ⇒ ـــــــ‬ = ‫ـــــــــ‬
∴ y = ‫ــــــــ‬
( ‫ـــــــــ‬ , ‫ـــــــــ‬ ) ‫ﻫﻮ‬ ‫اﻟﻤﺘﺠﻪ‬ ∴
π
6
a
x
a
π
6
x
5
x
5
3
2
5 3
2
y
a
π
6
y
5
1
2
y
5
5
2 5
2
5 3
2

r
A = x2
+ y2
‫ﺣﻴﺚ‬

r
A ‫ﻳﺴﺎوي‬

r
A(x,y) ‫ﻃﻮل‬ ‫ان‬ (1

cosθ =
x
r
A
, sinθ =
y
r
A
‫ﻧﺴﺘﺨﺪم‬

r
A(x,y) ‫اﺗﺠﺎه‬ ‫ﻻﻳﺠﺎد‬ (2

98
( 5 - 1 ) ‫تمرينات‬
/ 1‫س‬
ً‫ﻼ‬‫ﻛ‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮﺟﻬﻪ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ارﺳﻢ‬ ‫ﺛﻢ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫وإﺗﺠﺎه‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬
:‫ﻣﻨﻬﺎ‬
(1 , 3 ) (‫ﺟـ‬ ، (-3 , 0) (‫ب‬ ، ( -2 , 2) (‫أ‬
( 0 , -8) (‫ز‬ ، ( -3 , -3 ) (‫و‬ ، ( 3 , -1 ) (‫ﻫـ‬ ، (0 , 6 ) (‫د‬
/ 2‫س‬
:‫ﻳﻠﻲ‬ ‫ﻛﻤﺎ‬ ‫واﺗﺠﺎﻫﻪ‬ ‫ﻃﻮﻟﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺟﺪ‬
‫)أ‬ || B || = 2 ، Q = ‫ـــــ‬
‫)ب‬ || B || = 2 ، Q = ‫ـــــ‬
‫)ﺟـ‬ || B || = 4 ، Q = π
‫)د‬ || B || = 3 ، Q = ‫ـــــ‬
‫)ﻫـ‬ || B || = 4 ، Q = ‫ـــــ‬
π
4
π
6
π
3
π
2
2
3

99
� ‫ﺣﻘﻴﻘﻲ‬ ‫ﺑﻌﺪد‬ ‫وﺿﺮﺑﻬﺎ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﺟﻤﻊ‬ [5�4�
: ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﺟﻤﻊ‬ [5-4-1]
‫اﻟﻤﺘﺠﻪ‬ ‫وﻳﻜﻮن‬ ‫اﻵﺧﺮ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫إﻧﺘﻬﺎﺋﻪ‬ ‫ﻧﻘﻄﺔ‬ ‫وﻣﻦ‬ ‫أﺣﺪﻫﻤﺎ‬ ‫ﻧﺮﺳﻢ‬ ً‫ﺎ‬‫ﻫﻨﺪﺳﻴ‬ A ، B ‫ﻣﺜﻞ‬ ‫ﻣﺘﺠﻬﻴﻦ‬ ‫ﻟﺠﻤﻊ‬
‫اﻟﻤﺘﺠﻬﻴﻦ‬‫ﺟﻤﻊ‬‫ﺣﺎﺻﻞ‬‫ﻫﻮ‬‫اﻟﺜﺎﻧﻲ‬‫اﻟﻤﺘﺠﻪ‬‫اﻧﺘﻬﺎء‬‫ﺑﻨﻘﻄﺔ‬‫وﻳﻨﺘﻬﻲ‬‫اﻻول‬‫اﻟﻤﺘﺠﻪ‬‫ﺑﺪء‬‫ﺑﻨﻘﻄﺔ‬‫ﻳﺒﺘﺪئ‬‫اﻟﺬي‬
‫ﻗﻄﺮ‬ ‫اﻟﻤﺠﻤﻮع‬ ‫ﻳﻤﺜﻞ‬ ‫إذ‬ ،‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﺑﻄﺮﻳﻘﺔ‬ ‫ﻣﺘﺠﻬﻴﻦ‬ ‫ﻣﺠﻤﻮع‬ ‫إﻳﺠﺎد‬ ‫وﻳﺘﻢ‬ (5-9) ‫ﺷﻜﻞ‬ ‫ﻻﺣﻆ‬
. ( 5 - 10 ) ‫ﺷﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫ﻓﻴﻪ‬ ‫ﻣﺘﺠﺎورﻳﻦ‬ ‫ﺿﻠﻌﻴﻦ‬ ‫اﻟﻤﺘﺠﻬﺎن‬ ‫ﻳﻜﻮن‬ ‫اﻟﺬي‬ ‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬
(5-10) ‫ﺷﻜﻞ‬ (5-9) ‫ﺷﻜﻞ‬
A ، ‫اﻟﻤﺘﺠﻬﺎن‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫واﺣﺪة‬ ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ ‫أﻧﻬﻤﺎ‬ ‫ﻳﻘﺎل‬ ‫ﻋﻨﺪﺋﺬ‬ ‫واﺣﺪ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺘﺠﻬﺎن‬ ‫ﻳﻘﻊ‬ ‫ﻗﺪ‬
. ( 5 - 11 ) ‫اﻟﺸﻜـــﻞ‬ ‫ﻓــــﻲ‬ ‫ﻛﻤـــﺎ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫ﻣﺘﻀـــﺎدان‬ A ، B ‫اﻟﻤﺘﺠﻬﺎن‬ ‫ﺑﻴﻨﻤﺎ‬ C
(5-11) ‫اﻟﺸﻜﻞ‬
( x , y )
c
O
B (-x , -y)
X
Y
A
A
A
B
A
A
B
B
A
+ B
B
+

100
‫اﻻﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫وﻣﺘﻌﺎﻛﺴﻴﻦ‬ ‫اﻟﻄﻮل‬ ‫ﻓﻲ‬ ‫ﻣﺘﺴﺎوﻳﻴﻦ‬ ‫وﻛﺎﻧﺎ‬ ‫واﺣﺪة‬ ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ A ، B‫اﻟﻤﺘﺠﻬﺎن‬ ‫ﻛﺎن‬ ‫ﻓﺎذا‬
‫ﻓﺎن‬ ‫وﻛﺎن‬
‫ﻻﺣﻆ‬
. -A ‫ﺑﺎﻟﺮﻣﺰ‬ A ‫اﻟﻤﺘﺠﻪ‬ ‫ﻟﺴﺎﻟﺐ‬ ‫ﻳﺮﻣﺰ‬
(5-5) ‫ﺗﻌﺮﻳﻒ‬
:‫ﻓﺎن‬ ‫ﻛﺎن‬ ‫إذا‬
A + B = (x1
, y1
) + (x2
, y2
) = (x1
+ x2
, y1
+ y2
)
� � 5 ‫ﻣﺜﺎل‬
A + B ‫ﻓﺠﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
A + B = (3 ,1) + (1 , 4) = (4 , 5)
(5 - 12) ‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ً‫ﺎ‬‫ﻫﻨﺪﺳﻴ‬ ‫ذﻟﻚ‬ ‫ﺗﻮﺿﻴﺢ‬ ‫وﻳﻤﻜﻦ‬
A ، B ‫ﻟﻠﻤﺘﺠﻬﻴﻦ‬ ‫اﻟﻤﻜﻤﻞ‬ ‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﻗﻄﺮ‬ ‫ﻳﻤﺜﻞ‬ A + B ‫أن‬ ‫ﻻﺣﻆ‬
|| A || = || B || = x2
+y2
A = (x, y)B = (-x , -y)
A = (x1
,y1
) ، B = (x2
, y2
)
A = (3 , 1) ، B = (1, 4)
Y
X
(4,5)
B(1,4)
A(3,1)
O(0,0)
B
A
A+B
(5 - 12) ‫اﻟﺸﻜﻞ‬

101
� � 6 ‫ﻣﺜﺎل‬
. A + B ‫ﻓﺄوﺟﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
A+ B = (-4, 3) + ( 5, -2) = (1 ,1)
� ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﺟﻤﻊ‬ ‫ﺧﻮاص‬ [5�4�2�
. ً‫ﺎ‬‫أﻳﻀ‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ A + B ‫ﻓﺎن‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ A ، B ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻛﺎن‬ ‫إذا‬ :‫اﻻﻧﻐﻼق‬ (1)
ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ ، A ، B ، C ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻛﺎن‬ ‫إذا‬ :‫اﻟﺘﺠﻤﻴﻊ‬ (2)
(A + B) + C = A + (B +C) ‫ﻓﺎن‬
A + B = B +A ‫ﻓﺎن‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ A ، B ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻛﺎن‬ ‫إذا‬ :‫اﻟﺘﺒﺪﻳﻞ‬ (3)
‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻓﻲ‬ ‫اﻟﺠﻤﻊ‬ ‫ﻟﻌﻤﻠﻴﺔ‬ ‫اﻟﻤﺤﺎﻳﺪ‬ ‫اﻟﻌﻨﺼﺮ‬ ‫ﻫﻮ‬ ‫اﻟﺼﻔﺮي‬ ‫اﻟﻤﺘﺠﻪ‬ :‫اﻟﺠﻤﻌﻲ‬ ‫اﻟﻤﺤﺎﻳﺪ‬ ‫وﺟﻮد‬ (4)
A+ (0,0) = (0,0) + A = A ‫ﻓﺎن‬ ‫ﻣﺘﺠﻪ‬ ‫أي‬ A ‫ﻛﺎن‬ ‫إذا‬ :‫وﻣﻌﻨﺎه‬
B =- ‫ﻫﻮ‬ ‫آﺧﺮ‬ ‫ﻣﺘﺠﻪ‬ ‫ﻓﻴﻮﺟﺪ‬ ‫ﻣﺘﺠﻪ‬ ‫أي‬ A ‫ﻛﺎن‬ ‫إذا‬ :‫اﻟﺠﻤﻌﻲ‬ ‫اﻟﻨﻈﻴﺮ‬ ‫وﺟﻮد‬ (5)
‫ﺑﺤﻴﺚ‬
B = C ‫ﻓﺎن‬ A + B = A+ C ‫وﻛﺎن‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ A ، B ‫ﻛﺎن‬ ‫اذا‬ :‫اﻟﺤﺬف‬ ‫ﺧﺎﺻﻴﺔ‬ (6)
� � 7 ‫ﻣﺜﺎل‬
(-2, 3) ‫ﻟﻠﻤﺘﺠﻪ‬ ‫اﻟﺠﻤﻌﻲ‬ ‫اﻟﻨﻈﻴﺮ‬ ‫ﺟﺪ‬
: ‫(ﻷن‬ 2 , -3) ‫ﻫﻮ‬ (-2 , 3) ‫ﻟﻠﻤﺘﺠﻪ‬ ‫اﻟﺠﻤﻌﻲ‬ ‫اﻟﻨﻈﻴﺮ‬
( -2 , 3 ) + (2 , -3) = (-2 + 2, 3+( -3))= (0 , 0)
A + (B) = (B) + A = (0,0)
A= (-4 , 3) ، B = ( 5, -2)
A

102
�‫ﺣﻘﻴﻘﻲ‬ ‫ﺑﻌﺪد‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﺿﺮب‬ [5�4�3�
(5 - 6)‫ﺗﻌﺮﻳﻒ‬
‫ﺗﻮﺿﻴﺢ‬ ‫وﻳﻤﻜﻦ‬ ‫ﻓﺈن‬ ‫ﺣﻘﻴﻘﻲ‬ ‫ﻋﺪد‬ ‫أي‬ K ‫وﻛﺎن‬ ‫ﻛﺎن‬ ‫إذا‬
: ‫ﻳﻠﻲ‬ ‫ﻛﻤﺎ‬ ً‫ﺎ‬‫ﻫﻨﺪﺳﻴ‬ ‫اﻟﺘﻌﺮﻳﻒ‬ ‫ﻫﺬا‬
K ‫أي‬ || A || K ‫ﻳﺴﺎوي‬ ‫وﻃﻮﻟﻪ‬ A ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ ‫ﻳﻤﺜﻞ‬ KA ‫ﻓﺈن‬ ‫أن‬ ‫ﻧﻔﺮض‬
. ‫ﻧﻔﺴﻪ‬ A ‫اﻟﻤﺘﺠﻪ‬ ‫اﺗﺠﺎه‬ ‫وﻟﻪ‬ K > 0 ‫ﻳﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬ A ‫اﻟﻤﺘﺠﻪ‬ ‫ﻃﻮل‬ ‫ﺑﻘﺪر‬ ‫ﻣﺮة‬
‫وﻃﻮﻟﻪ‬ A ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ ‫ﻳﻘﻊ‬ K A ‫اﻟﻤﺘﺠﻪ‬ ‫ﻓﺎن‬ (‫)ﺳﺎﻟﺒﺔ‬ K < 0 ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫أﻣﺎ‬ (5 - 13) ‫اﻟﺸﻜﻞ‬ ‫ﻻﺣﻆ‬
. || A || K ‫ﻳﺴﺎوي‬
. A ‫ﻻﺗﺠﺎه‬ ‫ﻣﻌﺎﻛﺲ‬ ‫إﺗﺠﺎه‬ ‫وﻟﻪ‬ A ‫ﻃﻮل‬ ‫ﺑﻘﺪر‬ ‫ﻣﺮة‬ K ‫أي‬
(5 - 13) ‫ﺷﻜﻞ‬
A = (x , y)K A = A K = ( K x , K y )
A = ( x , y )
KyKx
Ky
X
Y
Kx
y
X
,Ky)(Kx
,y)(x
k > 0
k < 0
,
∨

103
� � 8 ‫ﻣﺜﺎل‬
2C ، C ، -3C ‫ﻓﺠﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
2C = 2(3,-1) = (6 , -2 )
C = (3,-1) = ( ‫ـــــ‬ , ‫ـــــ‬ )
-3C = -3 (3 , -1) = (-9,3)
� � 9 ‫ﻣﺜﺎل‬
K = 3 ، L = -2 ‫وﻛﺎن‬ ، ‫ﻛﺎن‬ ‫اذا‬
‫ﺟﺪ‬
(1) A+ B = ( 3 + 4 , -2 + 3) = (7 , 1)
(2) K A = 3 (3 , -2) = (9 , -6)
(3)L B = -2 (4 , 3) = ( -8 , -6 )
(4) K A + L B = (9 , - 6) + ( -8 , -6 )
= (1 , -12)
�‫ﺣﻘﻴﻘﻲ‬ ‫ﺑﻌﺪد‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﺿﺮب‬ ‫ﻋﻤﻠﻴﺔ‬ ‫ﺧﻮاص‬ [5�4�4�
: ‫ﻳﻜﻮن‬ ‫ﺣﻘﻴﻘﻲ‬ ‫ﻋﺪد‬ K ، ‫ﻣﺘﺠﻪ‬ A، B ‫ﻟﻜﻞ‬ :‫اﻟﺘﻮزﻳﻊ‬ ‫ﺧﺎﺻﻴﺔ‬ (1)
K ( A+ B) = K A+ K B
‫ﻛﺬﻟﻚ‬
:‫ﻳﻜﻮن‬ K ، L∈ R ‫ﻣﻦ‬ ‫وﻛﻞ‬ ‫ﻣﺘﺠﻪ‬ A ‫ﻟﻜﻞ‬ :‫اﻟﺘﺠﻤﻴﻊ‬ ‫ﺧﺎﺻﻴﺔ‬ (2)
(K × L) A = K ( L A) = L (K A)
C = (3, -1)1
2
3
2
1
2
-1
2
A = (3 , -2) ، B = (4 , 3)
(1) A + B (2) K A (3) L B (4) K A + L B
( A + B) K = A K + B K
1
2

104
1
0
‫ﺻﻔﺮ‬ ≠ K ‫ﺣﻴﺚ‬ K ∈R ، ‫ﻣﺘﺠﻪ‬ A ، B ‫ﻟﻜﻞ‬ :‫اﻟﺤﺬف‬ ‫ﺧﺎﺻﻴﺔ‬ (3)
. ‫وﺑﺎﻟﻌﻜﺲ‬ A= B ‫ﻓﺎن‬ K A = K B ‫ﻛﺎن‬ ‫ﻓﺎذا‬
× A = A × 1 = A (4)
× A = A × 0 = 0 (5)
� ‫ﻣﺘﺠﻬﻴﻦ‬ ‫ﻃﺮح‬ [5�4�5�
(5 - 7) ‫ﺗﻌﺮﻳﻒ‬
‫أﻧﻪ‬ ‫ف‬ّ‫ﺮ‬‫ﻳﻌ‬ A - B ‫ﻓﺎن‬ ً‫ﺎ‬‫ﻣﺘﺠﻬ‬ A ، B ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻛﺎن‬ ‫إذا‬
� � 10 ‫ﻣﺜﺎل‬
A - B ‫ﺟﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
A - B = A + (-B) = (3 , 4) + (1 , -3) = ( 4 , 1)
:‫ﻳﺄﺗﻲ‬ ‫ﻛﻤﺎ‬ ً‫ﺎ‬‫ﻫﻨﺪﺳﻴ‬ ‫ذﻟﻚ‬ ‫ﺗﻮﺿﻴﺢ‬ ‫وﻳﻤﻜﻦ‬
‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﻗﻄﺮ‬ ‫ﻳﻤﺜﻞ‬ A - B :‫أﻧﻪ‬ ‫أي‬
.B ‫اﻟﻤﺘﺠﻪ‬ ‫وﻟﺴﺎﻟﺐ‬ A ‫ﻟﻠﻤﺘﺠﻪ‬
(5 - 14) ‫اﻟﺸﻜﻞ‬
‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﻗﻄﺮ‬ ‫ﻳﻤﺜﻞ‬
.B ‫اﻟﻤﺘﺠﻪ‬ ‫وﻟﺴﺎﻟﺐ‬
A + (- B)
A = (3 , 4) ، B = (-1 , 3)
(3 ,4)
(4 ,1)
(1 ,-3)
(-1 ,3)
B A
A-B
B
X
Y
-

105
� � 11 ‫ﻣﺜﺎل‬
‫ﻛﺎن‬ ‫إذا‬
ً‫ﺎ‬‫ﻫﻨﺪﺳﻴ‬ ‫ذﻟﻚ‬ ‫ووﺿﺢ‬ KA - LB ( 2 ) A - B (1)
(1) A - B = (2 , 3)-(-2 , -1)
= (2 , 3)+(2 , 1)=(4 , 4)
: ‫ذﻟﻚ‬ ‫ﻳﻮﺿﺢ‬ (5- 15) ‫واﻟﺸﻜﻞ‬
( 5 - 15 ) ‫اﻟﺸﻜﻞ‬
(2) K A- L B = 2 (2 ,3) - (-1)(-2,-1)
= (4 , 6) + (-2,- 1)
= (2 , 5)
:‫اﻵﺗﻲ‬ ‫ﺑﺎﻟﺮﺳﻢ‬ ‫ذﻟﻚ‬ ‫وﻳﻮﺿﺢ‬
( 5 - 16 ) ‫اﻟﺸﻜﻞ‬
- B ‫وﻧﺠﺪ‬ B ‫ﻧﺮﺳﻢ‬ ‫ﺛﻢ‬ A ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫ﻃﻮﻟﻪ‬ ‫ﺑﻘﺪر‬ ‫ه‬ّ‫ﺪ‬‫ﻧﻤ‬ ‫ﺛﻢ‬ A ‫ﻧﺮﺳﻢ‬
‫ﻧﻌﻮد‬ ‫اﻧﻨﺎ‬ ‫أي‬ -1 × - B = B ‫ﺛﻢ‬
A+ B ‫ﻧﺠﻤﻊ‬ ‫ﺛﻢ‬ ‫ﺛﺎﻧﻴﺔ‬ B ‫اﻟﻰ‬
. ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺴﺆال‬ ‫ﻓﻲ‬ ‫ﻓﻌﻠﻨﺎ‬ ‫ﻛﻤﺎ‬
A = (2 , 3) ، B = (-2 , -1) ، K = 2 ، L = -1
2
2
Y
X
(4 ,4)
(2 ,3)
(2 ,1)
(-2 , -1)
A
B
(4 ,6)
(2 ,5)
Y
X
B
(-2 ,-1)
KA
KA
-LB

106
�‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫إﻋﻄﺎء‬ [5�5�
Unit Vector � ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻪ‬ [ 5 � 5 � 1 �
(5 - 8) ‫ﺗﻌﺮﻳﻒ‬
‫وﻃﻮﻟﻬﺎ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﺪاﻳﺘﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮﺟﻬﻪ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻫﻮ‬ U 1
‫اﻻﺳﺎﺳﻲ‬ ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻪ‬ (1)
. ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻫﻮ‬ ‫واﺗﺠﺎﻫﻬﺎ‬ ‫واﺣﺪة‬ ‫وﺣﺪة‬
‫وﻃﻮﻟﻬﺎ‬ ‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﺪاﻳﺘﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮﺟﻬﻪ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﻫﻮ‬ U2
‫اﻷﺳﺎﺳﻲ‬ ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻪ‬ (2)
. ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ ‫اﻟﺼﺎدات‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻫﻮ‬ ‫واﺗﺠﺎﻫﻬﺎ‬ ‫واﺣﺪه‬ ‫وﺣﺪة‬
:‫ﻓﺎن‬ ‫ﻛﺎن‬ ‫اذا‬
C = (x , 0) + (0 , y)
C = x (1 , 0) + y ( 0 , 1)
U 1
، U 2
‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﺪﻻﻟﺔ‬ C ‫اﻟﻤﺘﺠﻪ‬ ‫ﻳﻤﺜﻞ‬ ‫ﻫﺬا‬ C = x U 1
+ y U 2
: ‫ﻛﺎﻵﺗﻲ‬ U1
، U2
‫ﺑﺪﻻﻟﺔ‬ (0 , 6) ، (0 , -2) ، (-3 , 0) ، ( 9 , 0) ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻛﺘﺎﺑﺔ‬ ‫وﻳﻤﻜﻨﻨﺎ‬ ً‫ﻼ‬‫ﻓﻤﺜ‬
9U1
، (-3 , 0) = -3U1
، (0 ,-2) = -2U2
، (0 , 6) = 6 U 2
� � 12 ‫ﻣﺜﺎل‬
. ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫اﻟﻨﺎﺗﺞ‬ ‫ﻋﻦ‬ ‫وﻋﺒﺮ‬ A + B ‫ﺟﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
A + B = ( 4 , 7 ) + ( -5 , 3 ) = ( -1, 10 ) = -(1, 0 ) + 10(0 , 1) = - U1
+ 10 U 2
U 1
= (1 , 0 )
U 2
= (0 , 1)
C = (x , y)
A = (4 , 7) ، B = (-5 , 3)
(9,0)=

107
: ‫اﻵﺗﻴﺔ‬ ‫اﻻﻣﺜﻠﺔ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ U 1
، U 2
‫ﺑﺪﻻﻟﺔ‬ ‫ﻣﺘﺠﻪ‬ ‫أي‬ ‫ﻛﺘﺎﺑﺔ‬ ‫ﻳﻤﻜﻨﻨﺎ‬ ‫اﻷﺳﺎس‬ ‫ﻫﺬا‬ ‫وﻋﻠﻰ‬
(2, 5) = 2U1
+ 5U2
(-4, 2) = -4 U 1
+ 2U 2
( -2 , -3 ) = -2 U1
- 3U2
ً‫ﻼ‬‫ﻓﻤﺜ‬ ‫ﻳﻤﺜﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺮﺗﺐ‬ ‫اﻟﺰوج‬ ‫إﻳﺠﺎد‬ ‫ﻧﺴﺘﻄﻴﻊ‬ ‫ﻓﺎﻧﻨﺎ‬ ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﺼﻴﻐﺔ‬ ‫اﻟﻤﺘﺠﻪ‬ ‫ﻛﺘﺐ‬ ‫وإذا‬
‫ﻓﺎن‬ A = 4 U 1
+ 5U2
.‫وﻫﻜﺬا‬ ‫ﻓﺎن‬ B=- 2U1
+ 3U2
� � 13 ‫ﻣﺜﺎل‬
A+ B ‫ﺟﺪ‬ A = U1
- 3U2
، B = 2U1
+ U2
A+ B = ( U1
- 3U2
) + (2U1
+ U2
) = U1
(1+2) + U2
(-3+1)=3U1
- 2U2
� � 14 ‫ﻣﺜﺎل‬
‫ﺑﺪﻻﻟﺔ‬ ‫ﻋﻨﻪ‬ ‫ﻋﺒﺮ‬ ‫ﺛﻢ‬ K A - L B ‫ﺟﺪ‬ K= 2 ، L = 3 ‫وﻛﺎن‬ ‫وﻛﺎن‬ ‫ﻛﺎن‬ ‫إذا‬
. ‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬
K A - L B = 2 (5, -3) - 3 (-3, 4)
= (10,-6) + (9 , -12)
= (19 ,-18)
= 19 U1
- 18 U2
A = (4,5)
B = (-2 , 3)
A = (5 , -3)B = (-3,4)
=(3,-2)

108
( 5 - 2 ) ‫تمرينات‬
/ 1‫س‬
: ‫ﺑﺎﻟﺮﺳﻢ‬ ً‫ﺎ‬‫ﻣﻮﺿﺤ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ّ‫ﻛﻞ‬ ‫واﺗﺠﺎه‬ ‫ﻣﻘﺪار‬ ‫ﺟﺪ‬
(-2 , -2) ، (3, 0) ، 3 U 1
+ U2
، -U1
- 2U2
/ 2‫س‬
:‫ﻣﺎﻳﺄﺗﻲ‬ ‫ﺑﺴﻂ‬
4(1 ,-1) ، 2(1 , -1) ، -7(1,5) ، 3 (2,-1 )+ 4(-1 , 5)، 7(3U1
+ 2U2
) ، -4(2U1
-U2
/ 3‫س‬
:U1
، U2
‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﻮاﺳﻄﺔ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻋﻦ‬ ّ‫ﺮ‬‫ﻋﺒ‬
(-1 ,4) ، (-3 , -5) ، (0 , -1) ، ( 5 , 3)، (2 , 0) ، (2 , 3)
/ 4‫س‬
‫ﺑﺤﻴﺚ‬ ‫ﻣﺘﺠﻪ‬ ‫أي‬ A ‫وﻛﺎن‬ x ، y ∈ R ‫ﺣﻴﺚ‬ ‫ﻛﺎن‬ ‫إذا‬
‫أن‬ ‫ﻋﻠﻰ‬ ‫ﺑﺮﻫﻦ‬ A + E = E + A = A
/ 5‫س‬
A = -B ‫أن‬ ‫أﺛﺒﺖ‬ ‫ﻛﺎن‬ ‫إذا‬
/ 6‫س‬
A = ( 3 , 1) ، B = ( 2 , 3) ، K = 3 ، L = -2 ‫ﻛﺎن‬ ‫إذا‬
: ‫ﻳﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ً‫ﻼ‬‫ﻛ‬ ‫ﻓﺠﺪ‬
/ 7‫س‬
U1
، U2
‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﻮاﺳﻄﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻣﻦ‬ ‫ﻣﺘﺠﻪ‬ ‫ﻛﻞ‬ ‫ﻋﻦ‬ ‫ﺑﺎﻟﺘﻌﺒﻴﺮ‬ 6 ‫اﻟﺴﺆال‬ ‫ﺣﻞ‬
/ 8‫س‬
U1
، U2
‫اﻟﻮﺣﺪة‬ ‫ﻣﺘﺠﻬﻲ‬ ‫ﺑﻮاﺳﻄﺔ‬ ‫اﻵﺗﻴﺔ‬ ‫اﻟﻤﺘﺠﻬﺎت‬ ‫ﻋﻦ‬ ّ‫ﺮ‬‫ﻋﺒ‬
‫ـــــــــ‬ ‫وإﺗﺠﺎﻫﻪ‬ 10 ‫ﻃﻮﻟﻪ‬ ‫)ب(ﻣﺘﺠﻪ‬ ، ‫ـــــــــ‬ ‫وإﺗﺠﺎﻫﻪ‬ 3 ‫ﻃﻮﻟﻪ‬ ‫ﻣﺘﺠﻪ‬ ( ‫)أ‬
‫وإﺗﺠﺎﻫﻪ‬ ‫ـــــــــ‬ ‫ﻃﻮﻟﻪ‬ ‫ﻣﺘﺠﻪ‬ ( ‫د‬ ) ، ‫ـــــــــ‬ ‫وإﺗﺠﺎﻫﻪ‬ 5 ‫ﻃﻮﻟﻪ‬ ‫ﻣﺘﺠﻪ‬ (‫)ﺟـ‬
/ 9‫س‬
. 2A + 3x = 5B : ‫ﺑﺤﻴﺚ‬ x ‫ﺟﺪ‬ ‫ﻛﺎن‬ ‫إذا‬
E = (0 , 0)
K B ، L A ، A +B ، K A + B ،K A - B ، K A + L B
K A - L B ، K (A + B) ، (L + K) A ، (L + K) (A + B) ،
K (L A + K B) ، K L ( A - B)
A + B = B + A = (0,0)
A = (5 , 2) ، B = (2 ,-4)
E = (x , y)
π
3π
4
3
4
π
6
π
)

109109109
6 ‫ﺍﻻﺣﺪﺍﺛﻴﺔ‬ ‫ﺍﻟﻬﻨﺪﺳﺔ‬ : ‫ﺍﻟﺴﺎﺩﺱ‬ ‫ﺍﻟﻔﺼﻞ‬
. ‫االحداثي‬ ‫النظام‬ [6-1]
. ‫معلومتين‬ ‫نقطتين‬ ‫بين‬ ‫المسافة‬ [6-2]
. ( ‫الداﺧل‬ ‫من‬ ) ‫معلوم‬ ‫مستقيم‬ ‫تقسيم‬ ‫نقطة‬ ‫احداثيات‬ [6-3]
. ‫المستقيم‬ ‫ميل‬ [6-4]
� ‫اﻟﺘﻮازي‬ ‫ﺷﺮط‬ [6-5]
. ‫التعامد‬ ‫شرط‬ [6-6]
. ‫المستقيم‬ ‫معادلة‬ [6-7]
. ‫معلوم‬ ‫مستقيم‬ ‫عن‬ ‫معلومة‬ ‫نقطة‬ ‫عد‬ُ‫ب‬ [6-8]

n1
n2
n1
x2
+n2
x1
n1
+n2
n1
y2
+n2
y1
n1
+n2
|ax1
+ by1
+c|
a2
+b2
↔ ↔
↔ ↔
L = (x�
-x1
)2
+ (y2
_y1
)2
‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬
( , ) ‫ﻧﺴﺒﺘﻪ‬ ‫ﺗﻘﺴﻴﻢ‬ ‫ﻧﻘﻄﺔ‬ ‫اﺣﺪاﺛﻴﺎت‬
L1
// L2
m1
= m2
L1
,L2
‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺗﻮازي‬
L1
L2
m1
× m2
= -1 L1
,L2
‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺗﻌﺎﻣﺪ‬
ax + by + c = 0 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬
D = D ‫وﻣﺴﺘﻘﻴﻢ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬
:‫ان‬ ‫على‬ ً‫ا‬‫ر‬‫قاد‬ ‫الطالب‬ ‫يكون‬ ‫بان‬ ‫استه‬‫ر‬‫د‬ ‫في‬ ‫الفصل‬ ‫هذا‬ ‫يهدف‬
‫االحداثي‬ ‫النظام‬ ‫على‬ ‫يتعرف‬ -
‫االحداثي‬ ‫المستوى‬ ‫في‬ ‫نقطتين‬ ‫بين‬ ‫البعد‬ ‫يوجد‬ -
‫مستقيمة‬ ‫قطعة‬ ‫منتصﻒ‬ ‫احداثي‬ ‫يوجد‬ -
‫مستقيمة‬ ‫قطعة‬ ‫تقسيم‬ ‫نقطة‬ ‫احداثي‬ ‫يوجد‬ -
‫بمتﻐيرين‬ ‫االولى‬ ‫الدرجة‬ ‫من‬ ‫معادلة‬ ‫على‬ ‫يتعرف‬ -
‫المستقيم‬ ‫ميل‬ ‫على‬ ‫يتعرف‬ -
‫بمتﻐيرين‬ ‫االولى‬ ‫الدرجة‬ ‫من‬ ‫معادلة‬ ‫يوجد‬ -
‫ميولهما‬ ‫ﺧالل‬ ‫من‬ ‫والمتعامدان‬ ‫المتوازيان‬ ‫المستقيمان‬ ‫بين‬ ‫يميز‬ -
‫معلوم‬ ‫مستقيم‬ ‫عن‬ ‫معلومة‬ ‫نقطة‬ ‫عد‬ُ‫ب‬ ‫يوجد‬ -

110
Analgtic Geometrey : ‫اﻻﺣﺪاﺛﻴﺔ‬ ‫اﻟﻬﻨﺪﺳﺔ‬ : ‫اﻟﺴﺎدس‬ ‫اﻟﻔﺼﻞ‬
�‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫اﻻﺣﺪاﺛﻲ‬ ‫اﻟﻨﻈﺎم‬ [6�1�
‫وﻣﺜﻠﻨﺎ‬ (O) ‫ﻓﻲ‬ ‫وﻣﺘﻘﺎﻃﻌﻴﻦ‬ x xَ، y yَََ ‫ﻣﺘﻌﺎﻣﺪﻳﻦ‬ ‫ﻣﺴﺘﻘﻴﻤﻴﻦ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ‫رﺳﻤﻨﺎ‬ ‫إذا‬ ‫أﻧﻪ‬ ‫ﺗﻌﻠﻢ‬
‫ﻓﺎﻧﻨﺎ‬ ‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺗﻤﺜﻞ‬ O ‫أﻧﻪ‬ ‫واﻓﺘﺮﺿﻨﺎ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﻫﺬﻳﻦ‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻋﻠﻰ‬ (R) ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬
‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ y yَ، ‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ x xَ ‫وﻧﺴﻤﻲ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻓﻲ‬ ً‫ﺎ‬‫أﺣﺪاﺛﻴ‬ ً‫ﺎ‬‫ﻧﻈﺎﻣ‬ ‫أﻧﺸﺄﻧﺎ‬ ‫ﻗﺪ‬ ‫ﻧﻜﻮن‬ ‫ﺑﺬﻟﻚ‬
‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ ‫ﻋﻠﻰ‬ ‫اﻷول‬ ‫ﻋﻤﻮدﻳﻦ‬ ‫ﻣﻨﻬﺎ‬ ‫وﻧﺮﺳﻢ‬ A ‫ﻣﺜﻞ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫ﻧﻘﻄﺔ‬ ‫أﻳﺔ‬ ‫ﻧﺄﺧﺬ‬ ‫وﻋﻨﺪﻣﺎ‬
(6 - 1) ‫اﻟﺸﻜﻞ‬ ‫ﻻﺣﻆ‬ ‫اﻟﺘﺮﺗﻴﺐ‬ ‫ﻋﻠﻰ‬ A B ، A C ‫وﻟﻴﻜﻮﻧﺎ‬ ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻋﻠﻰ‬ ‫واﻵﺧﺮ‬
ً‫ﻻ‬‫أو‬ ‫اﻟﺴﻴﻨﻲ‬ ‫اﻻﺣﺪاﺛﻲ‬ ‫ﻳﺄﺗﻲ‬ ‫اﻟﺤﻘﻴﻘﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬ ‫ﻣﺮﺗﺐ‬ ‫زوج‬ ‫ﻋﻦ‬ ‫ﻋﺒﺎرة‬ ‫ﻧﻜﺘﺐ‬ ‫وﻋﻨﺪﻣﺎ‬
.‫اﻟﺼﺎدي‬ ‫اﻷﺣﺪاﺛﻲ‬ ‫ﺛﻢ‬
‫ﺗﺪرﻳﺞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺴﺘﺨﺪﻣﺔ‬ ‫اﻟﻄﻮل‬ ‫وﺣﺪة‬ ‫وأن‬ ‫ﻣﺘﻌﺎﻣﺪان‬ ‫اﻻﺣﺪاﺛﻴﺎت‬ ‫ﻣﺤﻮري‬ ‫ان‬ ‫ﺳﻨﻌﺘﺒﺮ‬ ‫اﻟﻔﺼﻞ‬ ‫ﻫﺬا‬ ‫ﻓﻲ‬
. ‫اﻵﺧﺮ‬ ‫اﻟﻤﺤﻮر‬ ‫ﺗﺪرﻳﺞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺴﺘﺨﺪﻣﺔ‬ ‫ﻧﻔﺴﻬﺎ‬ ‫ﻫﻲ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫أﺣﺪ‬
( 6 - 1) ‫ﺷﻜﻞ‬
A ( 3 , 2)
↔ ↔
↔
Y
X
A( 3,2)
C( 0 ,2)
O( 0 ,0) B( 3 ,0)
↔

111
�Distance Between Two Points ‫ﻣﻌﻠﻮﻣﺘﻴﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ [6 � 2�
‫ﺑﺎﻟﻄﺮﻳﻘﺔ‬ ‫إﻳﺠﺎدﻫﺎ‬ ‫ﻳﻤﻜﻦ‬ ‫ﺑﻴﻨﻬﻤﺎ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﻓﺎن‬ ‫اﻟﻤﺴﺘﻮي‬ ‫اﻟﻰ‬ ‫ﺗﻨﺘﻤﻴﺎن‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫إﺣﺪاﺛﻲ‬ ‫ﻋﺮﻓﻨﺎ‬ ‫إذا‬
:‫اﻵﺗﻴﺔ‬
:‫اﻟﻤﺴﺘﻮي‬‫ﻓﻲ‬‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻟﺘﻜﻦ‬
C ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ :A B C ∆
‫ﻓﻴﺜﺎﻏﻮرس‬ ............ L2
= +
L = ( x2
- x1
) 2
+(y2
- y1
)2
.‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﻗﺎﻧﻮن‬
: ‫أﺧﺮى‬ ‫ﺑﻄﺮﻳﻘﺔ‬ ‫أو‬
A B = B -A ‫اﻟﺨﺎﺻﻴﺔ‬ ‫ﺑﺎﺳﺘﺨﺪام‬
A B = (x2
, y2
) - ( x1
, y1
)
(6-2)‫ﺷﻜﻞ‬ = (x2
-x1
، y2
- y1
)
‫ﻧﻘﻄﺘﻴﻦ‬‫ﺑﻴﻦ‬‫اﻟﻤﺴﺎﻓﺔ‬‫..ﻗﺎﻧﻮن‬
� � 1 ‫ﻣﺜﺎل‬
.‫واﺣﺪ‬ ‫ﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﻨﻘﺎط‬ ‫أﻧﻪ‬ ‫أﺛﺒﺖ‬
: ‫اﻷوﻟﻰ‬ ‫اﻟﻄﺮﻳﻘﺔ‬
A B = B -A = (-3 , 4) - (-2, 7) = (-1 , -3)
AC = C -A = (1 , 16) - (-2 , 7) = (3 , 9) = -3 (-1 ,-3)
∴ A C = -3A B
.‫واﺣﺪ‬ ‫ﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ A، B ، C∴
:‫اﻟﻤﺴﺘﻮي‬‫ﻓﻲ‬‫ﻧﻘﻄﺘﻴﻦ‬
‫ﻓﻴﺜﺎﻏﻮرس‬ ............
A (-2 , 7) ، B (-3, 4) ، C (1, 16)
|| A B ||= ( x2
- x1
) 2
+(y2
- y1
)2
A (x1
, y1
) ، B ( x2
, y2
)
= (AC)2
( BC)2
B( x2
,y2
)
L
A( x1
,y1
)
y2
-y1
x2
-x1 C
O
y
x

112
: ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻄﺮﻳﻘﺔ‬
AB = ( -2 + 3 )2
+ ( 7 - 4 )2
= 1 + 9 = 10
BC = ( -3 - 1 )2
+ ( 4 - 16 )2
= 16 + 144 = 160 = 4 10
AC = ( -2-1)2
+ (7-16)2
= 9 +81 = 90 = 3 10
BC = AB + AC
‫أﻛﺒﺮ‬ ∆ ‫أي‬ ‫ﻓﻲ‬ ‫ﺿﻠﻌﻴﻦ‬ ‫أي‬ ‫ﻣﺠﻤﻮع‬ ‫أن‬ ‫إذ‬ ‫ﻣﺜﻠﺚ‬ ‫رؤوس‬ ‫ﻟﻜﺎﻧﺖ‬ ّ‫ﻻ‬‫وإ‬ ‫واﺣﺪ‬ ‫ﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ A، B، C
.‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﻀﻠﻊ‬ ‫ﻣﻦ‬
� � 2 ‫ﻣﺜﺎل‬
‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ‫ﻣﺜﻠﺚ‬ ‫ﻫﻮ‬ ‫اﻟﻨﻘﺎط‬ ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ∆ ‫اﻟـ‬ ‫أن‬ ‫ﺑﺮﻫﻦ‬
AB = ( 2- 1)2
+(2-1)2
= 1+1 = 2
AC = ( 5- 1)2
+(-1-1)2
= 16+4 = 20
BC = ( 5- 2)2
+(-1-2)2
= 9+9 = 18
( 20 )2
= ( 2 )2
+ ( 18 )2
: ‫ﻷن‬ AC2
= AB2
+ BC2
∵
20 = 2 + 18 ‫أن‬ ‫أي‬
. B ‫ﻓﻲ‬ ‫ﻗﺎﺋﻢ‬ ABC ∆ ∴
A (1 , 1) ، B (2 , 2) ، C (5 , -1)

113
� � 3 ‫ﻣﺜﺎل‬
.‫أﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫رؤوس‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﻨﻘﺎط‬ ‫أن‬ ‫ﺑﻴﻦ‬
AB = ( -3 -1 )2
+ ( -1 + 4 )2
= 16 + 9 = 25 =5
BC= ( 1 -10 )2
+ ( -4 + 5 )2
= 81 +1 = 82
CD = ( 10 - 6 )2
+ ( -5 + 2 )2
= 16 + 9 = 25 =5
AD = ( 6 + 3 )2
+ (-2 +1 )2
= 81 + 1 = 82
AB = CD ، BC = AD ‫أن‬ ‫وﺣﻴﺚ‬
.( ‫ﻣﺘﺴﺎوﻳﻴﻦ‬ ‫ﻣﺘﻘﺎﺑﻠﻴﻦ‬ ‫ﺿﻠﻌﻴﻦ‬ ‫ﻛﻞ‬ ‫ﻻن‬ ) ‫أﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﻳﻤﺜﻞ‬ ABCD ‫اﻟﺸﻜﻞ‬
� � 4 ‫ﻣﺜﺎل‬
‫ﻓﻴﻪ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬ ‫رؤوس‬ ‫ﻫﻲ‬ ‫اﻟﻨﻘﻂ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬
. a R ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬ AB=AC
AB=AC ‫ﻣﻌﻄﻰ‬
‫اﻟﻄﺮﻓﻴﻦ‬ ‫ﺑﺘﺮﺑﻴﻊ‬
‫ﻟﻠﻄﺮﻓﻴﻦ‬ ‫اﻟﺘﺮﺑﻴﻌﻲ‬ ‫اﻟﺠﺬر‬ ‫ﺑﺄﺧﺬ‬
( ‫ذﻟﻚ‬ ‫ﺳﺒﺐ‬ ‫)ﺑﻴﻦ‬ ‫ﺗﻬﻤﻞ‬
A(-3, -1) ، B (1 , -4) ، C (10, -5) ، D (6 , -2)
C(4, 1) ، B (a,1) ، A (3, 2a)
∈
B
A
C
⇒ 3− a( )
2
+ 2a −1( )
2
= 3− 4( )
2
+ 2a −1( )
2
⇒ 3−a( )
2
+ 2a −1( )
2
=1+ 2a −1( )
2
⇒ 3−a( )
2
=1
⇒ 3− a = ±1
:3−a =1⇒ a = 2
:3−a = −1⇒ a = 4‫او‬
‫اﻣﺎ‬
(3 , 2a)
(a,1) (4,1)

114
( 6 - 1 ) ‫تمرينات‬
/ 1‫س‬
:‫اﻷﺗﻴﺔ‬ ‫اﻟﻨﻘﺎط‬ ‫ﻣﻦ‬ ‫زوج‬ ‫ﻛﻞ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﺟﺪ‬
. (1 , 2) ، (6 , 4) (‫ب‬ ، .(0 , 0) ، (3 , 4) (‫أ‬
.(-2 , 3) ، (-1 , 4) (‫د‬ ، ( 5 , -1 ) ، ( -3 , - 5) (‫ﺟـ‬
/ 2‫س‬
.A(5 , 7) ، B( 1, 10) ، C (-3 , -8) ‫اﻟﻨﻘﺎط‬ ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺜﻠﺚ‬ ‫ﻣﺤﻴﻂ‬ ‫ﺟﺪ‬
/ 3‫س‬
.‫ﻗﻄﺮﻳﻪ‬ ‫ﻃﻮل‬ ‫ﺟﺪ‬ ‫ﻫﻲ‬ ‫رﺑﺎﻋﻲ‬ ‫ﺷﻜﻞ‬ ‫رؤوس‬
/ 4‫س‬
‫ﻣﺘﻮازي‬ ‫رؤوس‬ ‫ﻫﻲ‬ A (3 , -2) ، B ( -5 , 0) ، C (0 , -7) ، D (8 ,-9) ‫اﻟﻨﻘﺎط‬ ‫أن‬ ‫أﺛﺒﺖ‬
.‫اﻻﺿﻼع‬
/ 5‫س‬
‫إﺣﺪاﺛﻲ‬ ‫ﺟﺪ‬ ABCD ‫اﻻﺿﻼع‬ ‫ﻣﺘﻮازي‬ ‫ﻣﻦ‬ ‫رؤوس‬ ‫ﺛﻼث‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬
.D ‫ﻧﻘﻄﺔ‬
/ 6‫س‬
.‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬ ‫ﻫﻮ‬ ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺜﻠﺚ‬ ‫أن‬ ‫ﺑﻴﻦ‬
/ 7‫س‬
.‫واﺣﺪة‬ ‫اﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻘﻊ‬ (-3 , -4) ، (6 , 8) ، (0 , 0) ‫اﻟﻨﻘﻂ‬ ‫ان‬ ‫أﺛﺒﺖ‬
A (2 , 3) ، B (-1 , -1) ، C (3 , -4)
A(-2 , 5) ،B (3 , 3) ،C (-4 , 2)
A(4 , -3) ، B( 7 , 10) ، C (-8 ,2) ، D (-1 , -5)

115
� (‫اﻟﺪاﺧﻞ‬ ‫)ﻣﻦ‬ ‫ﻣﻌﻠﻮم‬ ‫ﺗﻘﺴﻴﻢ‬ ‫ﻧﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ [6�3�
‫ﺑﺤﻴﺚ‬ ‫ﻧﻬﺎﻳﺘﻴﻬﺎ‬ ‫ﻧﻘﻄﺘﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺗﻘﻊ‬ ‫ﻧﻘﻄﺔ‬ ‫اﺣﺪاﺛﻴﺎت‬ ‫إﻳﺠﺎد‬ ‫اﻟﺪاﺧﻞ‬ ‫ﻣﻦ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻗﻄﻌﺔ‬ ‫ﺑﺘﻘﺴﻴﻢ‬ ‫ﻳﻘﺼﺪ‬
. ‫ﻣﻌﻠﻮﻣﺔ‬ ‫ﺑﻨﺴﺒﺔ‬ ‫ﺗﻘﺴﻤﻬﺎ‬
‫أن‬ ‫وﻟﻨﻔﺮض‬
‫اﻟﺪاﺧﻞ‬ ‫ﻣﻦ‬ A B ‫ﺗﻘﺴﻢ‬ ‫اﻟﺘﻲ‬ C ‫إﻳﺠﺎد‬ ‫واﻟﻤﻄﻠﻮب‬
‫ﻧﻔﺮض‬ : ‫ﻧﻘﻮل‬ ‫ﻟﺬﻟﻚ‬ n1
: n2
‫ﺑﻨﺴﺒﺔ‬
‫ـــــــــ‬ = ‫ـــــــــ‬
X = ‫ـــــــــــــــــــــــــ‬ ‫ﻛﺬﻟﻚ‬ Y = ‫ـــــــــــــــــــــــــ‬ ‫ﻓﺎن‬
( ‫ـــــــــــــــــــــــــ‬ , ‫ـــــــــــــــــــــــــ‬ ) C ‫اﻟﺘﻘﺴﻴﻢ‬ ‫ﻧﻘﻄﺔ‬
� � 4 ‫ﻣﺜﺎل‬
‫اﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻮاﺻﻠﺔ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻗﻄﻌﺔ‬ ‫ﺗﻘﺴﻢ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
‫ـــــــــ‬ ‫ﺑﻨﺴﺒﺔ‬
x = ‫ـــــــــــــ‬ = ‫ـــــــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــــــ‬ = 1
y = ‫ـــــــــــــ‬ = ‫ـــــــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــــــ‬ = -2
( 1 , -2) ‫ﻫﻲ‬ ‫اﻟﺘﻘﺴﻴﻢ‬ ‫ﻧﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ∴
‫اﻟﺪاﺧﻞ‬ ‫ﻣﻦ‬
X = ‫ـــــــــــــــــــــــــ‬ ‫ﻛﺬﻟﻚ‬X = ‫ـــــــــــــــــــــــــ‬ ‫ﻛﺬﻟﻚ‬X
A = (x1
, y1
) ، B = (x2
, y2
)
C = (x , y)
n1
x2
+n2
x1
n1
+n2
n1
y2
+n2
y1
n1
+n2
n1
n2
AC
CB
n1
x2
+n2
x1
n1
+n2
n1
y2
+n2
y1
n1
+n2
1
2
A (4 , -3 ) ، B (-5, 0)
1 (-5)+2(4)
1+2
n1
x2
+n2
x1
n1
+n2
n1
y2
+n2
y1
n1
+n2
1 (0)+2(-3)
1+2
-6
3
-5+8
3
B(x 2
,y 2
)
C(x ,y)
A(x 1
,y 1
)
n2
n1

116
: ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﺗﻨﺼﻴﻒ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻧﺘﻴﺠـﺔ‬
AB‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﺗﻨﺼﻴﻒ‬ ‫ﻧﻘﻄﺔ‬ M ‫ﻧﻔﺮض‬
‫ﺣﻴﺚ‬
‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻧﻘﻄﺔ‬ M = ( ‫ـــــــــــــــــــــــــ‬ , ‫ـــــــــــــــــــــــــ‬ ) ‫ﻓﺎن‬
. ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﻓﻲ‬ ‫وﻋﻮض‬ n1
= n2
= n ‫إﺟﻌﻞ‬ ‫وﻟﻼﺛﺒﺎت‬
� � 5 ‫ﻣﺜﺎل‬
‫ﺣﻴﺚ‬ AB ‫ﻣﻨﺘﺼﻒ‬ C ‫ﻛﺎﻧﺖ‬ ‫إذا‬
C ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
C = ( ‫ـــــــــــــــــــ‬ , ‫ـــــــــــــــــــ‬ )
= ( ‫ـــــــــــــــــــ‬ , ‫ـــــــــــــــــــ‬ )
C = ( 2 , -3 )
x1
+x2
2
A(-3 , 2) ، B (7 , -8)
A ( x1
, y1
) ، B ( (x 2
, y2
)
y1
+y2
2
x1
+x2
2
y1
+y2
2
2 + (-8)
2
-3+7
2
‫ﺑﻨﺴﺒﺔ‬ ‫اﻟﺪاﺧﻞ‬ ‫ﻣﻦ‬ B(x2
,y2
) A(x1
,y1
) ‫ﺑﻴﻦ‬ ‫اﻟﻮاﺻﻠﺔ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﺗﻘﺴﻢ‬ C(x,y) ‫اﻟﻨﻘﻄﺔ‬
:‫ﻫﻲ‬ n1
n2
(
x1 + x2
2
,
y1 + y2
2
) :‫ﻫﻲ‬ ‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻧﻘﻄﺔ‬ ‫اﺣﺪاﺛﻴﺎت‬
C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)
x2
C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)C(
n1 + n2x1
n1 + n2
,
n1y2 + n2y1
n1 + n2
)

117
A (2 , 1) ، B (1, -3)
( 6 - 2 ) ‫تمرينات‬
/ 1‫س‬
‫ﺣﻴﺚ‬ ، A B ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﺗﻘﺴﻢ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
. ‫ـــــــــ‬ ‫ﺑﻨﺴﺒﺔ‬
/ 2‫س‬
. ‫أن‬ ‫ﺣﻴﺚ‬ AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻗﻄﻌﺔ‬ ‫ﺗﻨﺼﻒ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
/ 3‫س‬
‫أن‬ ‫ﺣﻴﺚ‬ ‫ـــــــــ‬ ‫ﺑﻨﺴﺒﺔ‬ AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻗﻄﻌﺔ‬ ‫ﺗﻘﺴﻢ‬ ‫اﻟﺘﻲ‬ C ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
/ 4‫س‬
، ‫ﺣﻴﺚ‬ B ‫ﻋﻦ‬ ‫ﺑﻌﺪﻫــــــﺎ‬ ‫أﻣﺜــــﺎل‬ ‫ﺛﻼﺛﺔ‬ A ‫ﻋﻦ‬ ‫ﺗﺒﻌﺪ‬ ‫اﻟﺘﻲ‬ C ‫اﻟﻨﻘﻄﺔ‬ ‫إﺣﺪاﺛﻴــــﺎت‬ ‫ﺟﺪ‬
B (4 , -4)
/ 5‫س‬
‫أﻃﻮال‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ : ‫ﺣﻴﺚ‬ A B C∆‫أﺿﻼع‬ ‫ﻣﻨﺘﺼﻔﺎت‬ ‫إﺣﺪاﺛﻴﺎت‬ ‫ﺟﺪ‬
.‫اﻟﻤﻘﺎﺑﻠﺔ‬ ‫اﻻﺿﻼع‬ ‫وﻣﻨﺘﺼﻔﺎت‬ ‫اﻟﻤﺜﻠﺚ‬ ‫رؤوس‬ ‫ﺑﻴﻦ‬ ‫اﻟﻮاﺻﻠﻪ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺎت‬
/ 6‫س‬
‫ﻳﻨﺼﻒ‬ (-1 ,-2) ، (1 , 3) ، ( -3 , -3 ) ، (-5 , -8) ‫رؤوﺳﺔ‬ ‫اﻟﺬي‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬ ‫ﻗﻄﺮي‬ ‫أن‬ ‫ﺑﻴﻦ‬
.‫اﻵﺧﺮ‬ ‫أﺣﺪﻫﻤﺎ‬
A(4 , 0) ، B (5 , 2) ، C (2 , -3)
A (2 , -4) ، B (-3 , -6)
A (1 , 3) ، B (4 , 6)
A(2 , 6)
2
1
3
5

118
� Slope of The Line ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻞ‬ْ‫ﻴ‬َ‫ﻣ‬ [6 � 4�
(6 - 1) ‫ﺗﻌﺮﻳﻒ‬
‫ﻓﺎن‬ B (x2
, y2
) ، A ( x1
, y1
) ‫ﻛﺎﻧﺖ‬ ‫إذا‬
. x1
≠ x2
‫ﺑﺸﺮط‬ ‫ـــــــــــــــــــ‬ = AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬
ً‫ا‬‫ﺮ‬‫ﺻﻔ‬ = AB ‫ﻣﻴﻞ‬ ‫أن‬ ‫ﻳﻌﻨﻲ‬ y2
- y1
= 0 ‫ﻛﺎن‬ ‫إذا‬ (1
. ‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ // AB ‫أن‬ ‫أي‬
.‫ﺻﻔﺮ‬ = ُ‫ﻪ‬‫ﻟ‬ ٍ‫ز‬‫ﻣﻮا‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻛﻞ‬ ‫ﻣﻴﻞ‬ = ‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ ‫ﻣﻴﻞ‬ ‫أن‬ ‫ﺑﻤﻌﻨﻰ‬
‫ﻣﻌﺮف‬ ‫ﻏﻴﺮ‬ AB ‫ﻣﻴﻞ‬ ‫أن‬ ‫ﻳﻌﻨﻲ‬ x2
- x1
= 0 ‫ﻛﺎن‬ ‫إذا‬ (2
.‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ // AB ‫أن‬ ‫أي‬
. ‫ﻣﻌﺮف‬ ‫ﻏﻴﺮ‬ ‫وﻳﻜﻮن‬ ‫ﻟﻪ‬ ً‫ﺎ‬‫ﻣﻮازﻳ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻛﻞ‬ ‫ﻣﻴﻞ‬ = ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻣﻴﻞ‬ ‫أن‬ ‫ﺑﻤﻌﻨﻰ‬
‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻷﺗﺠﺎه‬ ‫ﻣﻊ‬ AB ‫ﻳﺼﻨﻌﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮﺟﺒﺔ‬ ‫اوﻳﺔ‬‫ﺰ‬‫ﻟﻠ‬ ً‫ﺎ‬‫ﻗﻴﺎﺳ‬ Q ‫ﻛﺎﻧﺖ‬ ‫إذا‬ (3
. Q ∈ [ 0 , 180 ْ ) / { 90ْ } ‫ﺣﻴﺚ‬ tanQ ‫ﻳﺴﺎوي‬ ‫ﻣﻴﻞ‬ ‫ﻓﺎن‬ ‫اﻟﺴﻴﻨﺎت‬
� � 6 ‫ﻣﺜﺎل‬
A (2 , 3) ، B (5 , 1)‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ‫ﺟﺪ‬
m AB = ‫ــــــــــــــــ‬ = ‫ــــــــــــــ‬ = ‫ـــــــــــــ‬
y2
- y1
x2
- x1
y2
- y1
x2
- x1
1 - 3
5 - 2
- 2
3
AB

119
� Parallel Condition ‫اﻟﺘﻮازي‬ ‫ﺷﺮط‬ [6 � 5�
. m1
= m2
‫اذا‬ ‫وﻓﻘﻂ‬ ‫اذا‬ L1
// L2
‫ي‬ ‫ا‬ ‫وﺑﺎﻟﻌﻜﺲ‬ ‫ﻧﻔﺴﻪ‬ ‫اﻟﻤﻴﻞ‬ ‫ﻟﻬﻤﺎ‬ ‫اﻟﻤﺘﻮازﻳﺎن‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺎن‬
� � 7 ‫ﻣﺜﺎل‬
. ‫واﺣﺪ‬ ‫ﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ A(4 , 3) ، B (2 , 1) ، C (1 , 0 )‫اﻟﻨﻘﺎط‬ ‫ان‬ ‫ﺑﻴﻦ‬
m AB = ‫ـــــــــــ‬ = ‫ــــــــــــ‬ =
m BC = ‫ـــــــــــ‬ = ‫ــــــــــــ‬ =
m AB = mBC ∵
. ‫واﺣﺪ‬ ‫ﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ C ، B ، A ∴
� Perpendicular Condition ‫اﻟﺘﻌﺎﻣﺪ‬ ‫ﺷﺮط‬ [6 � 6�
‫اذا‬ ‫وﻓﻘﻂ‬ ‫اذا‬ L1
⊥ L2
‫أي‬ ‫وﺑﺎﻟﻌﻜﺲ‬ -1 = ‫ﻣﻴﻼﻫﻤﺎ‬ ‫ﺿﺮب‬ ‫ﺣﺎﺻﻞ‬ ‫ﻓﺎن‬ ‫ﻣﺴﺘﻘﻴﻤﺎن‬ ‫ﺗﻌﺎﻣﺪ‬ ‫إذا‬
m1
× m2
= -1
‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬‫ﻛﺎن‬‫إذا‬ً‫ﻼ‬‫ﺜ‬َ‫ﻣ‬‫اﻻﺷﺎرة‬‫ﺑﻌﻜﺲ‬‫اﻵﺧﺮ‬‫ﻣﻘﻠﻮب‬‫ﻳﺴﺎوي‬‫أﺣﺪﻫﻤﺎ‬‫ﻣﻴﻞ‬‫ي‬‫ا‬ m1
= ‫ـــــــــــ‬ ‫او‬
‫ﻣﻴﻠﻪ‬ ‫ﻳﻜـﻮن‬ ‫ﻋﻠﻴﻪ‬ ‫ﻋﻤﻮد‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫وأي‬ ‫ـــــــــ‬ = ‫ﻣﻴﻠﻪ‬ ‫ﻳﻜﻮن‬ ‫ﻳﻮازﻳﻪ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻓﺎي‬ ‫ــــــــــ‬ ‫ﻳﺴﺎوي‬
. ‫ـــــــــــ‬ =
1
1
- 2
- 2
- 1
- 1
1 - 3
2 - 4
0 - 3
1 - 2
- 1
m2
- 3
44
- 3
3
4

120
� � 8 ‫ﻣﺜﺎل‬
‫ﻗﺎﺋﻢ‬ ‫ﻫﻮ‬ A (3 , -1) ، B (10, 4) ، C (5 , 11) ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺜﻠﺚ‬ ‫أن‬ ‫اﻟﻤﻴﻞ‬ ‫ﺑﺎﺳﺘﺨﺪام‬ ‫ﺑﺮﻫﻦ‬
‫؟‬ B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬
mAB = ‫ـــــــــــــ‬ = ‫ـــــــ‬ , mBC = ‫ــــــــــــ‬ = ‫ــــــــــــ‬
mAB × mBC = ‫ـــــــــــ‬ × ‫ــــــــــــ‬ = -1 ∴
AB ⊥ BC ∴
. B ‫ﻓﻲ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ∆ ABC ∴
� � 9 ‫ﻣﺜﺎل‬
‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬ ‫واﺣﺪة‬ ‫اﺳﺘﻘﺎﻣﻪ‬ ‫ﻋﻠﻰ‬ C (-2 , b-4) ، B (-1, 2) ، A (0 , b) ‫اﻟﻨﻘﻂ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬
. b R
A,B,C ‫واﺣﺪة‬ ‫اﺳﺘﻘﺎﻣﻪ‬ ‫ﻋﻠﻰ‬
4 - (-1)
5 - 10
5
710 - 3
11 - 4 7
-5
7
7
5
-5
∈

⇒ mAB = mBC
⇒
2 −b
−1− o
=
b− 4( )− 2
−2 −1
2 −b
−1
=
b−6
−1
⇒ 2 −b = b−6
⇒ b= 4
m=
Vy
Vx

⇒ mAB = mBC
⇒
2 −b
−1−o
=
b− 4( )−2
−2 −1
2 −b
−1
=
b−6
−1
⇒ 2 −b= b−6
⇒ b= 4
m=
Vy
Vx
∴

121
( 6 - 3 ) ‫تمرينات‬
/ 1‫س‬
.(0 , -2) ، (2 , 0 ) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ‫ﺟﺪ‬ (1)
.‫واﺣﺪة‬ ‫إﺳﺘﻘﺎﻣﺔ‬ ‫ﻋﻠﻰ‬ (2 , 3) , (-1 , 4) ، ( -7 , 6) ، ‫اﻟﻨﻘﺎط‬ ‫أن‬ ‫ﺑﻴﻦ‬ (2)
AB = ‫ـــــــــ‬ ‫ﻳﻜﻮن‬ ‫ﺑﺤﻴﺚ‬ h ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬ A(2 , 3) ، B ( -3 , h )‫ﻛﺎﻧﺖ‬ ‫إذا‬ (3)
A(1 , 6) ، B (-2 , -8) ، C (7 , -2) ‫رؤوﺳﺔ‬ ‫ﻣﺜﻠﺚ‬ ABC (4)
. B ‫ﻣﻦ‬ ‫اﻟﻤﺎر‬ ABC ‫ﻟﻠﻤﺜﻠﺚ‬ ‫اﻟﻤﺘﻮﺳﻂ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ‫ﺟﺪ‬
/ 2‫س‬
: ً‫ة‬‫ﻓﻘﺮ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻻﺟﺎﺑﺔ‬ ‫ﺣﺪد‬ ، ‫ﺻﺤﻴﺤﺔ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻓﻘﻂ‬ ‫واﺣﺪة‬ ‫إﺟﺎﺑﺎت‬ ‫أرﺑﻊ‬ ‫ﻳﺄﺗﻲ‬ ‫ﻓﻴﻤﺎ‬ ً‫ة‬‫ﻓﻘﺮ‬ ‫ﻟﻜﻞ‬
‫ﻳﺴﺎوي‬ L ‫ﻣﻴﻞ‬ ‫ﻓﺎن‬ (1 , 5) ، (2 , 3) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻳﻤﺮ‬ H ، L ⊥ H ‫ﻛﺎن‬ ‫إذا‬ (1
(‫د‬ ، ( ‫ﺟـ‬ ، -2 (‫ب‬ ، (‫أ‬
‫ﻳﺴﺎوي‬ L ‫ﻣﻴﻞ‬ ‫ﻓﺎن‬ ( -3 , 2) ، (3 , -2) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻳﻤﺮ‬ H ، L ⁄ ⁄ H ‫ﻛﺎن‬ ‫إذا‬ (2
(‫د‬ ، ( ‫ﺟـ‬ ، - (‫ب‬ ، (‫أ‬
‫ﻓﺎن‬ (3 , 4) ، (x , 6) ∈ H ، (-1 , 3) ، (-1 , 5) ∈ L ، ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻛﺎن‬ ‫اذا‬ (3
.‫ﺻﺤﻴﺢ‬ ‫ﺳﺒﻖ‬ ‫ﻣﻤﺎ‬ ً‫ﺎ‬‫اﻳ‬ ‫ﻟﻴﺲ‬ (‫د‬ 1 ( ‫ﺟـ‬ ، 3 (‫ب‬ ، -3 (‫أ‬ ‫ﻳﺴﺎوي‬ x ‫ﻗﻴﻤﺔ‬
1
2
1
2
2
33
- 2
3
2
2
33
- 2 -3
2
L ⁄ ⁄ H
m

122
/ 3‫س‬
.‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻗﺎﺋﻢ‬ ∆ ‫رؤوس‬ ‫ﻫﻲ‬ A(5, 2) ، B (-2 , 1)، C (2 , -2) ‫اﻟﻨﻘﺎط‬ ‫ان‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﻴﻞ‬ ‫ﺑﺎﺳﺘﺨﺪام‬ (1
.‫اﺿﻼع‬ ‫ﻣﺘﻮازي‬ ABCD ‫اﻟﺸﻜﻞ‬ ‫ان‬ ‫ﺑﻴﻦ‬ A(-1, 5)، B (5 , 1)، C (6 ,-2)، D(0 , 2)‫ﻟﺘﻜﻦ‬ (2
.‫ﻣﺮﺑﻊ‬ ABCD ‫اﻟﺸﻜﻞ‬ ‫ان‬ ‫ﺑﻴﻦ‬ A(5 , 2)، B (2 ,-1)، C (-1 , 2)، D (2 , 5) ‫ﻟﺘﻜﻦ‬ (3
:‫ﺟﺪ‬ A (2 , 4) ، B (6 , 0)، C (-2 , -3) ‫رؤوﺳﻪ‬ ‫ﻣﺜﻠﺚ‬ ABC (4
. BC ‫ﻋﻠﻰ‬ A ‫ﻣﻦ‬ ‫اﻟﻤﺮﺳﻮم‬ ‫اﻟﻌﻤﻮد‬ ‫ﻣﻴﻞ‬ ( ‫أ‬
. AC ً‫ﺎ‬‫وﻣﻮازﻳ‬ B ‫ﻣﻦ‬ ‫اﻟﻤﺮﺳﻮم‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ (‫ب‬
‫ﻳﻤﺜﻞ‬ A (-2 , 2) ، B (2 , -2) ، C (4 , 2) ، D (2 , 4 ) ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ‫اﻟﺮﺑﺎﻋﻲ‬ ‫اﻟﺸﻜﻞ‬ ‫ان‬ ‫ﺑﻴﻦ‬ (5
.‫اﻟﻘﻄﺮﻳﻦ‬ ‫ﻣﺘﻌﺎﻣﺪ‬ ‫ﻣﻨﺤﺮف‬ ‫ﺷﺒﻪ‬
‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ً‫ا‬‫ﻋﻤﻮد‬ (x , 4)، (-2 , -9) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﺠﻌﻞ‬ ‫اﻟﺘﻲ‬ x ‫ﻗﻴﻤﺔ‬ ‫ﺟﺪ‬ (6
.(4 , 1) ، (0 , 3) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬

123
� Equation of The Line ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ [6�7�
‫ذﻟﻚ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﺴﻤﻰ‬ x ، y ‫ﺑﻴﻦ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﻓﺎن‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫أي‬ ‫ﻧﻘﺎط‬ ‫ﻣﻦ‬ ‫ﻧﻘﻄﺔ‬ ‫اﻳﺔ‬ ( x , y) ‫ﻛﺎﻧﺖ‬ ‫اذا‬
. ‫اﻟﻤﺴﺘﻘﻴﻢ‬
a x + b y + c = 0 :‫ﻫﻲ‬ ‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫اﻟﻌﺎﻣﺔ‬ ‫اﻟﻘﻴﺎﺳﻴﺔ‬ ‫واﻟﻤﻌﺎدﻟﺔ‬
x = 0 ‫ﺑﻮﺿﻊ‬ ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ‫ﺗﻤﺜﻴﻠﻪ‬ ‫ﻳﻤﻜﻦ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫ﻳﻘﻄﻊ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ .1
y = ‫ـــــــــ‬ ∴
x = ‫ ـــــــــ‬y = 0 ‫ﺑﻮﺿﻊ‬
‫وﻣﻨﻬﺎ‬ ‫اﻟﺼﺎدي‬ ‫اﻟﻤﺤﻮر‬ ‫ﻳﻮازي‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﻤﺜﻞ‬ ax + c = 0 ‫ﻳﻜﻮن‬ b = 0 ‫ﻳﻜﻮن‬ ‫وﻋﻨﺪﻣﺎ‬ .2
.‫اﻟﺼﺎدي‬ ‫اﻟﻤﺤﻮر‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﻤﺜﻞ‬ x = 0
‫وﻣﻨﻬﺎ‬ ‫اﻟﺴﻴﻨﻲ‬ ‫اﻟﻤﺤﻮر‬ ‫ﻳﻮازي‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﻤﺜﻞ‬ b y + c = 0 ‫ﻳﻜﻮن‬ a = 0 ‫ﻳﻜﻮن‬ ‫وﻋﻨﺪﻣﺎ‬ .3
.‫اﻟﺴﻴﻨﻲ‬ ‫اﻟﻤﺤﻮر‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﻤﺜﻞ‬ y = 0
.‫اﻷﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﻳﻤﺮ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗﻤﺜﻞ‬ ax + by = 0 ‫ﻳﻜﻮن‬ c = 0 ‫ﻳﻜﻮن‬ ‫وﻋﻨﺪﻣﺎ‬ .4
� ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻳﺠﺎد‬ ‫ﻛﻴﻔﻴﺔ‬
:‫ﻧﻘﻄﺘﺎن‬ ‫ﻣﻨﻪ‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ .1
: A (x1
, y1
) ، B (x2
, y2
) ‫ﺣﻴﺚ‬ AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬
:‫ﻓﺎن‬ C (x , y ) ∈ AB ‫ﻟﺘﻜﻦ‬
.‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫إﻳﺠﺎد‬ ‫ﻗﺎﻧﻮن‬ ‫ـــــــــــــــــــ‬ = ‫ــــــــــــــــــــ‬
:‫وﻣﻴﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﻣﻨﻪ‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ .2
m = ‫ـــــــــــــــــ‬ ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﻣﻦ‬
.‫وﻣﻴﻞ‬‫ﻧﻘﻄﺔ‬‫ﺑﺪﻻﻟﺔ‬‫اﻟﻤﺴﺘﻘﻴﻢ‬‫ﻣﻌﺎدﻟﺔ‬‫إﻳﺠﺎد‬‫ﻗﺎﻧﻮن‬ ............. y – y1
= m (x – x1
)
-c
b
-c
a
y - y1
x - x1
y2
- y1
x2
- x1
y2
- y1
x2
- x1
↔

124
� � 9 ‫ﻣﺜﺎل‬
.(2 , -3) ، (4 , 5) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
‫ــــــــــــــــ‬ = ‫ــــــــــــــــ‬
‫ــــــــــــــــ‬ = ‫ــــــــــــــــ‬
‫ـــــــــــــــ‬ = ‫ـــــــــــــــ‬
= 4x - 8
.‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ......... 4x – y – 11 = 0 ∴
� � 10 ‫ﻣﺜﺎل‬
‫ﻻ؟‬ ‫ام‬ ‫اﻟﻴﻪ‬ ‫ﺗﻨﺘﻤﻲ‬ (3 , 4) ‫اﻟﻨﻘﻄﺔ‬ ‫ان‬ ‫وﻫﻞ‬ ،(7,1) ،(0,3) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
‫ــــــــــــ‬ = ‫ـــــــــــ‬
2x -14 = -7y + 7
.‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ....... 2x + 7y – 21 = 0
. ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ x = 3 ، y = 4 ‫ﻋﻦ‬ ‫ﻧﻌﻮض‬ ،‫ﻻ‬ ‫ام‬ ‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻨﺘﻤﻲ‬ (3, 4) ‫ان‬ ‫ﻧﺘﺄﻛﺪ‬ ‫ﻟﻜﻲ‬
2( 3) + 7( 4 ) - 21 = 0
6 + 28 - 21 = 0
y + 3
x - 2
y - y1
x - x1
y2
- y1
x2
- x1
4 - 2
5 + 3
1
4y + 3
x - 2
3 - 1
0 - 7
y - 1
x - 7
y+3
?
?
(4,5 )
Y
X
(2, -3)
(3,4)
Y
X

125
13 ≠ 0 ∴
. ‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﻻﺗﻨﺘﻤﻲ‬ (3, 4) ‫اﻟﻨﻘﻄﺔ‬ ∴
� � 11 ‫ﻣﺜﺎل‬
. ‫ـــــــــ‬ ‫وﻣﻴﻠﻪ‬ (1 , -3) ‫اﻟﻨﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
y - y1
= m (x - x1
)
y + 3 = (x - 1 )
2y + 6 = x - 1
.‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ . . . . . . . . x - 2 y - 7 = 0
� � 12 ‫ﻣﺜﺎل‬
‫اﻟﺘﻲ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫اﻟﻘﻄﻌﺔ‬ ‫ﺗﻨﺼﻴﻒ‬ ‫وﻧﻘﻄﺔ‬ A(-2 , 5 )‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫ﻳﻤﺮ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
. B (4, -1)، C (-2 ,3 ) ‫اﻟﻨﻘﻄﺘﺎن‬ ‫ﻧﻬﺎﻳﺘﺎﻫﺎ‬
D = ( ‫ــــــــــــــ‬ , ‫ـــــــــــــــــ‬ ) = ⇐ B C ‫ﻣﻨﺘﺼﻒ‬ D ‫ﻟﺘﻜﻦ‬
‫ـــــــــــــــــ‬ = ‫ـــــــــــــــــ‬ : ‫ﻫﻲ‬ AD ‫ﻣﻌﺎدﻟﺔ‬
3y – 15 = -4x - 8 ∴
.‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ . . . . . . . . 4x + 3 y – 7 = 0
.‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ . . . . . . . .
D = ( ‫ــــــــــــــ‬ , ‫ـــــــــــــــــ‬ ) =
2
1
2
1
2
1 - 5
-1 + 3
2
4 + (-2) (1,1)
1 + 2
y - 5
x + 2
Y
X
(1 ,-3 )
D X
A
C
B
Y

126
� � 13 ‫ﻣﺜﺎل‬
. (-3, 5) ‫واﻟﻨﻘﻄﺔ‬ ‫اﻻﺻﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
. O (0 ,0) ، A (-3 ,5 )
‫ــــــــــــــ‬ = ‫ــــــــــــــ‬ : ‫ﻫﻲ‬ O A ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬
‫ــــــــــــ‬ = ‫ـــــــــــ‬
. ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ........ 5x + 3y = 0
:‫ﻣﻌﺎدﻟﺘﻪ‬ ‫ﻣﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ‫اﻳﺠﺎد‬ ‫ﻳﻤﻜﻦ‬ ‫ﻧﺘﻴﺠـﺔ‬
ax + by + c = 0 :‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ان‬ ‫ﻧﻔﺮض‬
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻣﻦ‬ ‫واﺣﺪ‬ ‫ﻃﺮف‬ ‫ﻓﻲ‬ x ، y ‫ﺑﺸﺮط‬ ‫اﻻﺷﺎرة‬ ‫ﺑﻌﻜﺲ‬ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬
b ≠ 0 ‫وان‬
‫ــــــــــــــ‬ - = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ∴
‫ــــــــــ‬ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ‫اي‬
. (-3, 5) ‫واﻟﻨﻘﻄﺔ‬ ‫اﻻﺻﻞ‬ ‫ﺑﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
‫ــــــــــــــ‬ = ‫ــــــــــــــ‬ : ‫ﻫﻲ‬y - 0
x - 0
5 - 0
-3 - 0
y 5
-3x
y ‫ﻣﻌﺎﻣﻞ‬
x ‫ﻣﻌﺎﻣﻞ‬
-a
b
x ‫-ﻣﻌﺎﻣﻞ‬
y ‫ﻣﻌﺎﻣﻞ‬
Y
X
A(-3,5)
O

127
� � 14 ‫ﻣﺜﺎل‬
:‫ﻣﻌﺎدﻟﺘﻪ‬ ‫اﻟﺬي‬ ‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫اﻟﺼﺎدي‬ ‫واﻟﻤﻘﻄﻊ‬ ‫اﻟﻤﻴﻞ‬ ‫ﺟﺪ‬
3x – 4 y – 12 = 0
m = ‫ــــــــــــ‬ = ‫ــــــــــــ‬ = ‫ــــــــــــ‬
:‫اﻟﺼﺎدي‬ ‫اﻟﻤﻘﻄﻊ‬
� � 15 ‫ﻣﺜﺎل‬
‫وﻳﻤﺮ‬ 150 ˚ ‫ﻗﻴﺎﺳﻬﺎ‬ ‫اوﻳﺔ‬‫ز‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻦ‬ ‫ﻳﺼﻨﻊ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
. (1 ,-4) ‫ﺑﺎﻟﻨﻘﻄﺔ‬
‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬
∵y - y1
= m (x - x1
)
∴ y + 4 = -1 (x - 1)
. ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ . . . . . . . . . x + 3 y + 4 3 – 1 = 0
-a -3 3
b -4 4
x = 0 ‫ﻋﻦ‬ ‫ﻧﻌﻮض‬
: ‫ﻓﻴﻜﻮن‬
m = tan 150˚
= tan (180˚-30˚)
= - tan 30˚
m = - 1
3
3
-4y - 12 =0 ⇒ y=-3

128
-a -2 2
b -3 3
-3
2
-3
2
� � 16 ‫ﻣﺜﺎل‬
‫ﻣﻌﺎدﻟﺘﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫وﻋﻤﻮدﻳ‬ (-2,1) ‫اﻟﻨﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﻳﻤﺮ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬
2x - 3y -7 = 0 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻦ‬ � ‫اﻟﺤــــﻞ‬
m = ‫ـــــــــ‬ = ‫ــــــــ‬ = ‫ـــــــــ‬ : ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ∴
. ( ‫ﻋﻠﻴﻪ‬ ً‫ﺎ‬‫ﻋﻤﻮدﻳ‬ ‫ﻻﻧﻪ‬ ) ‫ـــــــ‬ = ‫اﻟﻤﻄﻠﻮب‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ ∴
y – y1
= m (x – x1
)
y – 1 = ‫ـــــــ‬ (x + 2)
‫اﻟﻤﻄﻠﻮب‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ...... 3 x + 2 y + 4 = 0
2x - 3y -7 = 0
‫ﻫﻮ‬ ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ (1
‫ﻫﻮ‬ = 0 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫2(ﻣﻴﻞ‬
‫ﻫﻮ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻊ‬ ‫اوﻳﺔ‬‫ﺰ‬‫اﻟ‬ ‫ﻳﺼﻨﻊ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ (3
‫ﻫﻮ‬ ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ (4
‫ﻫﻮ‬ m ‫وﻣﻴﻞ‬ ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ (5
x2 ,y2( ), x1,y1( )m=
y2 -y1
x2 -x1
ax+by+c
θm = tanθ
y-y1
x-x1
=
y2 -y1
x2 -x1
x2 ,y2( ), x1,y1( )
x2 ,y2( ), x1,y1( )y-y1 =m x-x1( )
–m=
a
b
m=
a
b

129
( 6 - 4 ) ‫تمرينات‬
/ 1‫س‬
. (- 4 , 0 ) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫وﻳﻤﺮ‬ ‫ــــــــــــ‬ = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫1.ﺟﺪ‬
. (2 , -1) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫وﻳﻤﺮ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮازي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫2.ﺟﺪ‬
. ( 2 , -1) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫وﻳﻤﺮ‬ ‫اﻟﺼﺎدات‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮازي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫3.ﺟﺪ‬
. (-1, 5) ،(-1, 3) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫4.ﺟﺪ‬
. ‫ــــــــ‬ = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ L1
‫اﻟﻰ‬ ‫واﻟﻤﻮازي‬ (2 ,-1) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ L ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫5.ﺟﺪ‬
. ‫ــــــــ‬ = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫وﻋﻤﻮدﻳ‬ (0 ,-2) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫6.ﺟﺪ‬
(2, -2) ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫وﻋﻤﻮدﻳ‬ ( 3 , -4) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫7.ﺟﺪ‬
(0 ,3) ،
AB ‫ﻳﻨﺼﻒ‬ ‫اﻟﺬي‬ ‫اﻟﻌﻤﻮد‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬ ‫8.ﻟﺘﻜﻦ‬
/ 2‫س‬
‫وﺣﺪات‬ 7 ‫ﻃﻮﻟﻪ‬ ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ً‫ﺎ‬‫ﻣﻮﺟﺒ‬ ً‫ا‬‫ﺟﺰء‬ ‫وﻳﻘﻄﻊ‬ -3 = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫1.ﺟﺪ‬
‫وﺣﺪات‬ 6 ‫ﻃﻮﻟﻪ‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ً‫ﺎ‬‫ﺳﺎﻟﺒ‬ ً‫ا‬‫ﺟﺰء‬ ‫وﻳﻘﻄﻊ‬ 2 = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫2.ﺟﺪ‬
:‫ﻳﺎﺗﻲ‬ ‫ﻓﻴﻤﺎ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ‫اﻟﻤﻘﻄﻮع‬ ‫واﻟﺠﺰء‬ ‫اﻟﻤﻴﻞ‬ ‫3.ﺟﺪ‬
‫أ‬ . L1
: 2 x -3y + 5 = 0
‫ب‬ . L2
: 8 y = 4x + 16
‫ﺟـ‬ . L3
: 3 y = -4
: ‫ﻣﻌﺎدﻟﺘﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫وﻳﻮازي‬ ( 2 , -5) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫4.ﺟﺪ‬
2 x - y + 3 = 0
- 1
2
A (4,-2)، B (1, 2)
2
3
-3
5

130
ً‫ﺎ‬‫وﻋﻤﻮدﻳ‬ ‫وﺣﺪات‬ 4 ‫ﻃﻮﻟﻪ‬ ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ً‫ﺎ‬‫ﺳﺎﻟﺒ‬ ً‫ا‬‫ﺟﺰء‬ ‫ﻳﻘﻄﻊ‬ ‫اﻟﺬي‬ L ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫5.ﺟﺪ‬
. 2y = 4 x -1 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬
‫ﺛﻢ‬ ‫اﻟﺼﺎدات‬ ‫ﻣﺤﻮر‬ ‫ﻣﻊ‬ ‫ﺗﻘﺎﻃﻌﻪ‬ ‫وﻧﻘﻄﺔ‬ ‫ﻣﻴﻠﻪ‬ ‫ﺟﺪ‬ x + y -2= 0 :‫ﻣﻌﺎدﻟﺘﻪ‬ ً‫ﺎ‬‫ﻣﺴﺘﻘﻴﻤ‬ L ‫6.ﻟﻴﻜﻦ‬
L ‫ارﺳﻢ‬
‫ﻣﻌﺎدﻟﺘﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫وﻋﻤﻮدﻳ‬ (2, -2) ‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫اﻟﻤﺎر‬ L ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫7.ﺟﺪ‬
.‫اﻻﺣﺪاﺛﻴﻴﻦ‬ ‫اﻟﻤﺤﻮرﻳﻦ‬ ‫ﻣﻊ‬ L ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ x + y = 0
H : 3x + 6y = -3 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫و‬ L : 2x - y = 3 ‫8.اﻟﻤﺴﺘﻘﻴﻢ‬
L⊥ H ‫ان‬ ‫ﺑﻴﻦ‬ .‫أ‬
. L ، H ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬ ً‫ﺎ‬‫ﺟﺒﺮﻳ‬ ‫ﺟﺪ‬ .‫ب‬
‫ﺑﻨﻘﻄﺔ‬ ‫واﻟﻤﺎر‬ ‫اﻟﺴﻴﻨﺎت‬ ‫ﻟﻤﺤﻮر‬ ‫اﻟﻤﻮﺟﺐ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻣﻊ‬ 135 ˚ ‫ﻳﺼﻨﻊ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬ .9
.‫اﻻﺻﻞ‬
: ‫ﺟﺪ‬ (1, 2)‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫ﻳﻤﺮ‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ .10
a ∈ R
L2 : a +1( )×+y = 2
L1 :2y = ax+6
‫ﻗﻴﻤﺔ‬ ( ‫أ‬
L ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻴﻞ‬ (‫ب‬
‫اﻟﺼﺎدي‬ ‫ﻣﻘﻄﻌﻪ‬ (‫ﺟـ‬
L :2y = ax +1

131
‫ﻣﻌﻠﻮم‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫ﻣﻌﻠﻮﻣﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ [ 6 � 8 �
(6 - 2) ‫ﺗﻌﺮﻳﻒ‬
‫ﻋﻦ‬ N ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫ﻓﻴﻌﺮف‬ ‫ﻣﻌﻠﻮﻣﺔ‬ ‫واﻟﻨﻘﻄﺔ‬ L: ax +b y + c = 0 ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻛﺎن‬ ‫اذا‬
: ‫اﻻﺗﻴﺔ‬ ‫ﺑﺎﻟﻌﻼﻗﺔ‬ ‫وﺗﻌﻄﻰ‬ L ‫واﻟﻤﺴﺘﻘﻴﻢ‬ N ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻴﻦ‬ (D) ‫اﻟﻌﻤﻮدﻳﺔ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﺑﺎﻧﻪ‬ L ‫اﻟﻤﺴﺘﻘﻴﻢ‬
‫اﻟﺒﻌﺪ‬ ‫ﻗﺎﻧﻮن‬ . . . D = ‫ـــــــــــــــــــــــــــــــــ‬
� � 17 ‫ﻣﺜﺎل‬
. 2y + x = 2 : ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫ﻌﺪ‬ُ‫ﺑ‬ ‫ﺟﺪ‬
x + 2y – 2 =0 ⇐a = 1, b = 2, c = -2 : ‫اﻻﺗﻲ‬ ‫ﺑﺎﻟﺸﻜﻞ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﻀﻊ‬
= ‫ــــــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــــــــــــــــ‬
‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫اﻳﺠﺎد‬ ‫ﻳﻤﻜﻦ‬ ‫ﻧﺘﻴﺠـﺔ‬
: ‫ﺣﻴﺚ‬ L1
: a1
x + b1
y + c1
= 0، L2
:a 2
x + b2
y + c2
= 0
= L1
، L2
‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬
N (x1
,y1
)
A(1,3)
| a x1
+ b y1
+c |
a2
+ b2
| (1)(1)+(2)(3)-2|
(1)2
+(2)2
5
5
D = ‫ــــــــ‬ = 5 unit
| C2
-C1
|
a2
+b2
| a x1
+ b y1
+c |
a2
+ b2
L:ax+by+c=0
D=NM
N(x 1
,y 1
)
M
D
X
Y
0

132
� � 18 ‫ﻣﺜﺎل‬
:‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﺟﺪ‬
L1
: x - 3y = 1، L2
: x - 3y = 4
. ‫اﻵﺧﺮ‬ ‫ﻋﻦ‬ ‫ﻷﺣﺪﻫﻤﺎ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫ﻧﻘﻄﺔ‬ ‫أي‬ ‫ﺑﻌﺪ‬ ‫ﻫﻮ‬ ‫ﻣﺘﻮازﻳﻴﻦ‬ ‫ﻣﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬
L1
: y = 0 ⇒ x = 1 :‫ﻓﻲ‬ ‫ﻟﺬا‬
(1, 0) ‫اﻟﻨﻘﻄﺔ‬ ∴
D = ‫ـــــــــــــــــــــــــــــــ‬ ∴
D = ‫ـــــــــــــــــــــــــــــــ‬ =
: ‫اﻟﻨﺘﻴﺠﺔ‬ ‫ﺣﺴﺐ‬ ‫آﺧﺮ‬ ‫ﺣﻞ‬
D = ‫ـــــــــــــــ‬ =
:‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﺟﺪ‬
L
| (1)(1) - 3(0)- 4|
1 + 9
| a x1
+ b y1
+ c|
a2
+ b2
| 4-1 |
1 + 9
3
10
3
10
L1
L2
D

133
� � 19 ‫ﻣﺜﺎل‬
‫اﻟﻨﻘﺎط‬ ‫رؤوﺳﻪ‬ ‫اﻟﺬي‬ ‫اﻟـﻤﺜﻠﺚ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫ﺟﺪ‬
:AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫وﻟﻴﻜﻦ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫اﺿﻼع‬ ‫اﺣﺪ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻧﺠﺪ‬
‫ـــــــــــــــ‬ = ‫ـــــــــــــــ‬
‫ـــــــــ‬ = ‫ـــــــــــــــ‬ ⇒ ‫ــــــــــ‬ = ‫ـــــ‬
∴ 3 x – 2 y + 1 = 0
∆ ABC ‫ارﺗﻔﺎع‬ ‫ﻳﻤﺜﻞ‬ AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫اﻵن‬
‫ــــــــــــــــــــــــــــــ‬ = ‫ـــــــ‬
AB = (3-1)2
+ (5-2)2
= 4 + 9 = 13 ‫ﻃﻮل‬ ‫ﻧﺠﺪ‬
Area ∆ = ‫ـــــ‬ (AB) . D
= ‫ـــــــــ‬ ×( 13) . ‫ــــــ‬ = unit2
:AB ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫وﻟﻴﻜﻦ‬ ‫اﻟﻤﺜﻠﺚ‬ ‫اﺿﻼع‬ ‫اﺣﺪ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻧﺠﺪ‬
y2
- y1
x2
- x1
A (1,2) ، B (3 ,5) ، C ( 1 ,3)
y - y1
x - x1
C (-1,3)
D = 8
13
3
2
y - 2
x -1
5 - 2
3 -1
y - 2
x -1
| 3(-1) -2(3)+1 |
9 + 4
4
1
2
1
2
8
13
-
unit
X
Y
B
C D
A

134
( 6 - 5 ) ‫تمرينات‬
/ 1‫س‬
: ‫ﻳﺄﺗﻲ‬ ‫ﻓﻴﻤﺎ‬ ‫ﺧﺎﻃﺌﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ (×) ‫وﻋﻼﻣﺔ‬ ‫ﺻﺎﺋﺒﺔ‬ ‫اﻟﻌﺒﺎرة‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ (✓) ‫ﻋﻼﻣﺔ‬ ‫ﺿﻊ‬
.‫وﺣﺪات‬ 3 ‫ﻫﻮ‬ y = 3 :‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ .1
.‫وﺣﺪات‬ 5 ‫ﻫﻮ‬ y = -5 :‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ .2
.‫وﺣﺪات‬ 5 ‫ﻫﻮ‬ x = -5 :‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ .3
.‫وﺣﺪات‬ 3 ‫ﻫﻮ‬ y = 4، y = -1 :‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ .4
/ 2‫س‬
6x + 8y – 21 = 0 :‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ (-2 ,1) ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫1.ﺟﺪ‬
‫ﻣﺤﻮر‬ ‫ﻣﻦ‬ ً‫ﺎ‬‫ﻣﻮﺟﺒ‬ ً‫ا‬‫ﺟﺰء‬ ‫وﻳﻘﻄﻊ‬ ‫ـــــــــ‬ = ‫ﻣﻴﻠﻪ‬ ‫اﻟﺬي‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫اﻻﺻﻞ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫2.ﺟﺪ‬
.‫وﺣﺪات‬ 4 ‫ﻃﻮﻟﻪ‬ ‫اﻟﺼﺎدات‬
:‫اﻟﻤﺘﻮازﻳﻴﻦ‬ ‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫3.ﺟﺪ‬
L1
: 8x - 6y + 4 = 0
L2
: 4x - 3y - 1 = 0
. ‫ﺑﺎﻟﻨﻘﻄﺘﻴﻦ‬ ‫اﻟﻤﺎر‬ ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ (0 ,-2) ‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫4.ﺟﺪ‬
. ‫ﺣﻴﺚ‬ABC ‫اﻟﻤﺜﻠﺚ‬ ‫ﻣﺴﺎﺣﺔ‬ ‫5.ﺟﺪ‬
1
3
A(-4, 6)، B(-3, -1)، C (5, -2)
A (1,-1)، B (3, 5)

135135135
7 Statistics ‫ﺍﻹﺣﺼﺎﺀ‬ : ‫ﺍﻟﺴﺎﺑﻊ‬ ‫ﺍﻟﻔﺼﻞ‬
. ‫المركزية‬ ‫النزعة‬ ‫مقاييس‬ [7-1]
. ‫الحسابي‬ ‫الوسﻂ‬ [7-2]
. ‫الوسيﻂ‬ [7-3]
. ‫المنوال‬ [7-4]
� ‫اﻟﺘﺸﺘﺖ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ [7-5]

X ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
ME ‫اﻟﻮﺳﻴﻂ‬
MO ‫اﻟﻤﻨﻮال‬
R ‫اﻟﻤﺪى‬
S ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬
r ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬
:‫االحصاء‬ ‫تدريس‬ ‫اليها‬ ‫يهدف‬ ‫التي‬ ‫السلوكية‬ ‫االهداف‬ ‫اهم‬ ‫من‬ ‫ان‬ -
‫الحسابي‬ ‫الوسﻂ‬ ‫معنى‬ ‫على‬ ‫يتعرف‬ -
‫الحسابي‬ ‫الوسﻂ‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫الوسيﻂ‬ ‫على‬ ‫يتعرف‬ -
‫الوسيﻂ‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫المنوال‬ ‫على‬ ‫يتعرف‬ -
‫المنوال‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫المعياري‬ ‫اف‬‫ر‬‫االنح‬ ‫على‬ ‫يتعرف‬ -
‫المعياري‬ ‫اف‬‫ر‬‫االنح‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -
‫االرتباط‬ ‫معامل‬ ‫على‬ ‫يتعرف‬ -
‫االرتباط‬ ‫معامل‬ ‫ايجاد‬ ‫من‬ ‫يتمكن‬ -

136
: Statistics ‫اﻹﺣﺼﺎء‬ : ‫اﻟﺴﺎﺑﻊ‬ ‫اﻟﻔﺼﻞ‬
� Measures of Central Tendency ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اﻟﻨﺰﻋﺔ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ [ 7 � 1 �
‫ان‬ ‫ﻧﺮﻳﺪ‬ ‫واﻻن‬ ً‫ﺎ‬‫وﺑﻴﺎﻧﻴ‬ ً‫ﺎ‬‫ﺟﺪوﻟﻴ‬ ‫وﻋﺮﺿﻬﺎ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺟﻤﻊ‬ ‫اﺋﻖ‬‫ﺮ‬‫ﻃ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﺳﻴﺔ‬‫ر‬‫اﻟﺪ‬ ‫اﺣﻞ‬‫ﺮ‬‫اﻟﻤ‬ ‫ﻓﻲ‬ ‫اﺧﺬﻧﺎ‬
‫ﻋﻠﻰ‬ ‫اﻟﺤﺼﻮل‬ ‫ﻧﺮﻳﺪ‬ ‫أي‬ ،‫ﻟﻬﺎ‬ ً‫ﻼ‬‫وﻣﻤﺜ‬ ‫اﺳﺔ‬‫ر‬‫اﻟﺪ‬ ‫ﻣﻮﺿﻮع‬ ‫اﻟﻈﺎﻫﺮة‬ ‫ﻋﻦ‬ ً‫ا‬‫ﺮ‬‫ﻣﻌﺒ‬ ‫ﻳﻜﻮن‬ ‫ﻣﻘﻴﺎس‬ ‫ﻋﻦ‬ ‫ﻧﺒﺤﺚ‬
‫ﻫﺬا‬ ‫ﻓﻲ‬ ‫اﻟﺪﺧﻮل‬ ‫ﺟﻤﻴﻊ‬ ‫ﻋﻦ‬ ‫ﻳﻌﺒﺮ‬ ‫ﺑﻠﺪ‬ ‫ﻓﻲ‬ ً‫ﻼ‬‫ﻣﺜ‬ ‫اﻟﺪﺧﻞ‬ ‫ﻓﻤﺘﻮﺳﻂ‬ .‫اﻟﻘﻴﻢ‬ ‫ﺟﻤﻴﻊ‬ ‫ﻋﻦ‬ ‫ﺗﻌﺒﺮ‬ ‫واﺣﺪة‬ ‫ﻗﻴﻤﺔ‬
.‫ﻟﻠﺪﺧﻞ‬ ‫اﻟﻌﺎم‬ ‫اﻟﻤﺴﺘﻮى‬ ‫ﻋﻦ‬ ‫ﻳﻌﺒﺮ‬ ‫أي‬ ‫اﻟﺒﻠﺪ‬
‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻢ‬ ‫وﻫﺬه‬ ‫ﻣﺘﻮﺳﻄﺔ‬ ‫ﻣﻌﻴﻨﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺣﻮل‬ ‫ﺗﺘﺮﻛﺰ‬ ‫ﻻﻧﻬﺎ‬ ً‫ﻼ‬‫ﻣﻴ‬ ‫او‬ ‫ﻧﺰﻋﺔ‬ ‫ﻟﻬﺎ‬ ‫ان‬ ،‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺧﺼﺎﺋﺺ‬ ‫وﻣﻦ‬
.‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اﻟﻨﺰﻋﺔ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ ‫او‬ ‫ﺑﺎﻟﻤﺘﻮﺳﻄﺎت‬ ‫ﺗﺴﻤﻰ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺣﻮﻟﻬﺎ‬ ‫ﺗﺘﺮﻛﺰ‬
‫اﻟﻤﺮﺣﻠﺔ‬ ‫ﻓﻲ‬ ‫درﺳﺘﻬﺎ‬ ‫ان‬ ‫ﺑﻌﺪ‬ ‫اﻟﺘﻮﺳﻊ‬ ‫ﻣﻦ‬ ‫ﺑﺸﻲء‬ ‫اﻟﻤﺮﻛﺰﻳﺔ‬ ‫اﻟﻨﺰﻋﺔ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ ‫اﻫﻢ‬ ‫ﻧﺘﻨﺎول‬ ‫وﺳﻮف‬
:‫وﻫﻲ‬ ‫ﺑﺴﻴﻂ‬ ‫ﺑﺸﻜﻞ‬ ‫اﻟﻤﺘﻮﺳﻄﺔ‬
.‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ــ‬
.‫اﻟﻮﺳﻴﻂ‬ ‫ــ‬
.‫اﻟﻤﻨﻮال‬ ‫ــ‬
‫وﻋﻠﻴﻪ‬ ‫اﻳﺎه‬‫ﺰ‬‫ﻣ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻟﻜﻞ‬ ‫ﻛﻤﺎ‬ ‫اﻟﺤﺴﺎب‬ ‫وﻃﺮﻳﻘﺔ‬ ‫اﻟﻔﻜﺮة‬ ‫ﺣﻴﺚ‬ ‫ﻣﻦ‬ ‫اﻟﺜﻼﺛﺔ‬ ‫اﻟﻤﻘﺎﻳﻴﺲ‬ ‫ﻫﺬه‬ ‫وﺗﺨﺘﻠﻒ‬
.‫اﻻﺧﺮ‬ ‫دون‬ ‫ﻣﻦ‬ ‫اﻟﻤﻘﺎﻳﻴﺲ‬ ‫اﺣﺪ‬ ‫ﻓﻴﻬﺎ‬ ‫ﻳﺴﺘﺨﺪم‬ ‫اﻟﺘﻲ‬ ‫اﻟﺤﺎﻻت‬ ‫ﺑﻌﺾ‬ ‫وﺗﻮﺟﺪ‬ .‫ﻋﻴﻮﺑﻪ‬

137
� Arithmatic Mean ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ [ 7 � 2 �
(7 – 1 ) ‫ﺗﻌﺮﻳﻒ‬
‫ﻓﻲ‬ ‫ﻣﻔﺮدة‬ ‫ﻛﻞ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻣﺤﻞ‬ ‫ﺣﻠﺖ‬ ‫ﻟﻮ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺑﺎﻧﻪ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻳﻌﺮف‬
.‫اﻻﺻﻠﻴﺔ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻟﻤﺠﻤﻮع‬ ً‫ﺎ‬‫ﻣﺴﺎوﻳ‬ ‫اﻟﺠﺪﻳﺪة‬ ‫اﻟﻘﻴﻢ‬ ‫ﻫﺬه‬ ‫ﻣﺠﻤﻮع‬ ‫ﻟﻜﺎن‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬
. ‫ﻋﺪدﻫﺎ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻣﺠﻤﻮع‬ ‫ﻳﺴﺎوي‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻓﺎن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬
: ‫ﺣﺴﺎﺑﻪ‬ ‫ﻃﺮﻳﻘﺔ‬
‫اﻷوﻟﻰ‬ ‫اﻟﻄﺮﻳﻘﺔ‬
: ‫مبوبة‬ ‫ﻏير‬ (‫البيانات‬ ) ‫االحصائية‬ ‫المعلومات‬ ‫كانﺖ‬ ‫اذا‬ (1
‫ـــــــــــــــــــــــــــــــــ‬ = ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ : ‫وﺑﺎﻟﺮﻣﻮز‬
� � 1 ‫ﻣﺜﺎل‬
‫ﻫﺆﻻء‬ ‫ﻻﻋﻤﺎر‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫اﺣﺴﺐ‬ ‫ﺳﻨﺔ‬ 12,11,9,8,5 :‫ﻫﻲ‬ ‫اﺷﺨﺎص‬ ‫ﺧﻤﺴﺔ‬ ‫اﻋﻤﺎر‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬
.‫اﻻﺷﺨﺎص‬
‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ = ‫ــــــــــــــــــــــــــــــ‬
‫ـــــــــ‬ 9 ‫ﺳﻨﻮات‬
‫اﻟﻘﻴﻢ‬ ‫ﻣﺠﻤﻮع‬
‫ﻋﺪدﻫﺎ‬
x1
+ x2
+ x3
+ .......... + xn
n
X =
X =
x1
+ x2
+ x3
+ .......... + xn
n
12+ 11+ 9+ 8+5
5
45
5
=

138
: ‫مبوبة‬ ‫البيانات‬ ‫كانﺖ‬ ‫اذا‬ (2
:‫اﻻﺗﻲ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫اﺳﺘﺨﺪام‬ ‫ﻓﻴﻤﻜﻦ‬ ‫اري‬‫ﺮ‬‫ﺗﻜ‬ ‫ﺗﻮزﻳﻊ‬ ‫ﻓﻲ‬ ‫ﻣﺘﺠﻤﻌﺔ‬ ‫اﻻﺣﺼﺎﺋﻴﺔ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ = ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
� � 2 ‫ﻣﺜﺎل‬
(4) ‫و‬ ،‫ﺳﻨﻮات‬ 9 ‫ﻣﻨﻬﻢ‬ ‫ﻛﻞ‬ ‫ﻋﻤﺮ‬ ‫اﺷﺨﺎص‬ (5) ‫و‬ ،‫ﺳﻨﻮات‬ 8 ‫ﻣﻨﻬﻢ‬ ‫ﻛﻞ‬ ‫ﻋﻤﺮ‬ ‫اﺷﺨﺎص‬ (3) ‫وﺟﻮد‬ ‫ﻟﻨﻔﺮض‬
:‫اﻻﺗﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫ﺳﻨﺔ‬ 12 ‫ﻣﻨﻬﻢ‬ ‫ﻋﻤﺮﻛﻞ‬ ‫وﺷﺨﺼﻴﻦ‬ ،‫ﺳﻨﺔ‬ 11 ‫ﻣﻨﻬﻢ‬ ‫ﻛﻞ‬ ‫ﻋﻤﺮ‬ ‫اﺷﺨﺎص‬
‫اﻟﻮﺳﻂ‬ ‫اﺣﺴﺐ‬ ، ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﻳﻤﺜﻞ‬ ‫اﻟﺬي‬ ‫ﻫﻮ‬ (‫اﻟﻌﻤﺮ‬ ) ‫اﻟﻌﺪد‬ ‫ﻓﻴﻜﻮن‬ (‫ﻓﺌﺎت‬ ‫دون‬ ‫ﻣﻦ‬ ‫اﻟﺠﺪول‬ ‫ﻫﺬا‬ )
.‫ﻟﻠﻌﻤﺮ‬ ‫اﻟﺤﺴﺎﺑﻲ‬
‫اﻻﺷﺨﺎص‬ ‫وﻟﻌﺪد‬ x ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻠﻌﻤﺮ‬ ‫رﻣﺰﻧﺎ‬ ‫اذا‬
‫اﻟﺤـــــﻞ‬ ‫ﺧﻄﻮات‬ ‫ﻓﺎن‬ f ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫او‬
: ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫ﺗﺒﺴﻴﻄﻬﺎ‬ ‫ﻳﻤﻜﻦ‬
‫ــــــــــــــــــــــــ‬ ∴
= 9.786 ‫ﺳﻨﺔ‬
(‫ﻟﻠﻌﻤﺮ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫)اﻟﻮﺳﻂ‬
. ‫اﻟﻔﺌﺎت‬ ‫ذات‬ ‫ارﻳﺔ‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪاول‬ ‫ﺣﺎﻟﺔ‬ ‫وﻧﺎﺧﺬ‬ ‫اﺧﺮى‬ ‫ﺧﻄﻮة‬ ‫وﻟﻨﺘﻘﺪم‬
‫اﻟﻮﺳﻂ‬ ‫اﺣﺴﺐ‬ ، ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﻳﻤﺜﻞ‬ ‫اﻟﺬي‬ ‫ﻫﻮ‬ (‫اﻟﻌﻤﺮ‬ ) ‫اﻟﻌﺪد‬ ‫ﻓﻴﻜﻮن‬ (‫ﻓﺌﺎت‬ ‫دون‬ ‫ﻣﻦ‬ ‫اﻟﺠﺪول‬ ‫ﻫﺬا‬ )
121198‫اﻟﻌﻤﺮ‬
2453‫اﻻﺷﺨﺎص‬ ‫ﻋﺪد‬
‫اﻻﺷﺨﺎص‬ ‫وﻟﻌﺪد‬
‫اﻟﺤـــــﻞ‬ ‫ﺧﻄﻮات‬ ‫ﻓﺎن‬
‫اﻟﻌﻤﺮ‬ (x) ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ (f) ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ × ‫اﻟﻌﻤﺮ‬ (x f)
8 3 8 × 3 = 24
9 5 9 × 5 = 45
11 4 11 × 4 = 44
12 2 12 × 2 = 24
‫اﳌﺠﻤﻮع‬ 14 137
‫ارﻫﺎ‬‫ﺮ‬‫ﺗﻜ‬ ‫ﻓﻲ‬ ‫ﻓﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﻛﻞ‬ ‫ﺿﺮب‬ ‫ﺣﺎﺻﻞ‬ ‫ﻣﺠﻤﻮع‬
‫ات‬‫ر‬‫ا‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﻣﺠﻤﻮع‬
X =
x1
f1
+ x2
f2
+ x3
f3
+ .......... + xn
fn
f1
+f2
+........+fn
X =
137
14

139
� � 3 ‫ﻣﺜﺎل‬
‫اﻟﻮﺳﻂ‬ ‫ﺣﺴﺎب‬ ‫واﻟﻤﻄﻠﻮب‬ .‫ام‬‫ﺮ‬‫ﺑﺎﻟﻜﻴﻠﻮﻏ‬ ‫اﻟﻮزن‬ ‫ﻓﺌﺎت‬ ‫ﺣﺴﺐ‬ ‫ﺷﺨﺺ‬ ‫ﻣﺌﺔ‬ ‫ﺗﻮزﻳﻊ‬ ‫ﻳﺒﻴﻦ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬
‫؟‬ ‫ﻟﻠﻮزن‬ ‫اﻟﺤﺴﺎﺑﻲ‬
35 = ‫ــــــــــــــــــــــــ‬ = ‫اﻷوﻟﻰ‬ ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ : (x) ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﻧﺠﺪ‬
.‫وﻫﻜﺬا‬ ...........45 = 10 + 35 = ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬
: ‫ﻫﻲ‬ ‫اﻟﺤﻞ‬ ‫ﺧﻄﻮات‬ ‫ﻓﺎن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬
.(x) ‫ﻟﻬﺎ‬ ‫وﻧﺮﻣﺰ‬ ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫1(ﺣﺴﺎب‬
(f) ‫ارﻫﺎ‬‫ﺮ‬‫ﺗﻜ‬ ‫ﻓﻲ‬ (x) ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫2(ﻧﻀﺮب‬
:‫اﻟﻌﻼﻗﺔ‬ ‫ﻣﻦ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫3(ﻧﺠﺪ‬
61.1
:‫اﻟﻌﻼﻗﺔ‬ ‫ﻣﻦ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫(ﻧﺠﺪ‬
‫اﻟﻮزن‬ ‫ﻓﺌﺎت‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ (f) ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ (x) x × f
30- 9 35 315
40- 15 45 675
50- 22 55 1210
60- 25 65 1625
70- 18 75 1350
80- 90 11 85 935
‫اﳌﺠﻤﻮع‬ 100 6110
X =
X =
6110
100
30 + 40
2
‫ام‬‫ﺮ‬‫ﻏ‬ ‫ﻛﻴﻠﻮ‬
‫اﻟﻤﺠﻤﻮع‬80-9070-60-50-40-30-‫اﻟﻮزن‬ ‫ﻓﺌﺎت‬
10011182522159‫اﻻﺷﺨﺎص‬ ‫ﻋﺪد‬
x1
f1
+ x2
f2
+ x3
f3
+ .......... + xn
fn
f1
+f2
+........+fn
X =

140
� � 4 ‫ﻣﺜﺎل‬
: ‫اﻻﺗﻲ‬ ‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﺟﺪ‬
‫ــــــــــــــــــــــــــ‬
13.3
‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻄﺮﻳﻘﺔ‬
: ‫افات‬‫ر‬‫االنح‬ ‫أو‬ ‫الفرضي‬ ‫الوسﻂ‬ ‫ﻃريقة‬
‫إﻳﺠﺎد‬ ‫ﺛﻢ‬ ً‫ﺎ‬‫ﻓﺮﺿﻴ‬ ً‫ﺎ‬‫وﺳﻄ‬ ‫ﺑﻮﺻﻔﻬﺎ‬ ( ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ) ‫اﻟﻘﻴﻢ‬ ‫إﺣﺪى‬ ‫إﺧﺘﻴﺎر‬ ‫ﻋﻠﻰ‬ ‫اﻟﻄﺮﻳﻘﺔ‬ ‫ﻫﺬه‬ ‫ﺗﻌﺘﻤﺪ‬
: ‫اﻟﻘﺎﻧﻮن‬ ‫ﻧﻄﺒﻖ‬ ‫ﺛﻢ‬ ‫وﻣﻦ‬ ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ذﻟﻚ‬ ‫ﻋﻦ‬ ‫ﻓﺌﺔ‬ ‫ﻛﻞ‬ ‫اف‬‫ﺮ‬‫إﻧﺤ‬
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ + ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ = ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
X 0
= ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﺣﻴﺚ‬ X 0
+ ‫ـــــــــــــــــ‬
. ‫ات‬‫ر‬‫ا‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﻣﺠﻤﻮع‬ = ∑f ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ = X-X0
, ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ = f
‫اﻟﻤﺠﻤﻮع‬18-2016-14-12-10-8-‫اﻟﻔﺌﺎت‬
60461020155‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬
‫اﻟﻔﺌﺎت‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ (f) ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ (x) x × f
8- 5 9 45
10- 15 11 165
12- 20 13 260
14- 10 15 150
16- 6 17 102
18- 20 4 19 76
X = 798
60
X =
(‫ارﻫﺎ‬‫ﺮ‬‫ﺗﻜ‬ ‫ﻓﻲ‬ ‫ﻓﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫اف‬‫ﺮ‬‫)اﻧﺤ‬ ‫ﻣﺠﻤﻮع‬
‫ات‬‫ر‬‫ا‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﻣﺠﻤﻮع‬
∑ f. E
∑ f
X =
E

141
� � 5 ‫ﻣﺜﺎل‬
‫ﺑﻄﺮﻳﻘﺔ‬ ‫ﻟﻸﻋﻤﺎر‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫أوﺟﺪ‬ .‫ﺟﺎﻣﻌﻲ‬ ‫ﻃﺎﻟﺐ‬ 100 ‫أﻋﻤﺎر‬ ‫ﻳﺒﻴﻦ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪول‬
‫اﻟﻔﺮﺿﻲ؟‬ ‫اﻟﻮﺳﻂ‬
.‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﻧﺴﺘﺨﺮج‬ (1
. ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫أﻛﺒﺮ‬ ‫ﻳﻘﺎﺑﻞ‬ ‫اﻟﺬي‬ (21) ‫وﻟﻴﻜﻦ‬ ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ ( ) ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻧﺨﺘﺎر‬ (2
‫اﻟﻮﺳﻂ‬ - ‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬ = ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ) ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻋﻦ‬ ‫ﻓﺌﺔ‬ ‫ﻛﻞ‬ ‫ﻣﺮﻛﺰ‬ ‫اف‬‫ﺮ‬‫أﻧﺤ‬ ‫ﻧﺴﺘﺨﺮج‬ (3
.(‫اﻟﻔﺮﺿﻲ‬
. ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻋﻦ‬ ‫ﻣﺮﻛﺰﻫﺎ‬ ‫اف‬‫ﺮ‬‫اﻧﺤ‬ × (f) ‫ﻓﺌﺔ‬ ‫ﻛﻞ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫ﺿﺮب‬ ‫ﺣﺎﺻﻞ‬ ‫ﻧﺴﺘﺨﺮج‬ (4
‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﻤﻌﻠﻮﻣﺎت‬ ‫ﻧﻜﺘﺐ‬ ،∑ (f.E) ‫اﻟﻜﻠﻲ‬ ‫واﻟﻤﺠﻤﻮع‬ ‫ات‬‫ر‬‫ا‬‫ﺮ‬‫ﻟﻠﺘﻜ‬ ‫اﻟﻜﻠﻲ‬ ‫اﻟﻤﺠﻤﻮع‬ ‫ﻧﺴﺘﺨﺮج‬ (5
:‫ﻛﺎﻵﺗﻲ‬ ‫ﺟﺪول‬ ‫ﻓﻲ‬
‫ﻟﻼﻋﻤﺎر‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
‫اﻻﻋﻤﺎر‬
‫ﻟﻠﻔﺌﺎت‬
‫اﻟﻄﻼب‬ ‫ﻋﺪد‬
‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ (f)
‫اﻟﻔﺌﺔ‬ ‫ﻣﺮﻛﺰ‬
(X)
‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬
f.E
18- 20 19 19 - 21 = -2 20 × -2 = -40
20- 44 21=X0
21 - 21 = 0 44 × 0 = 0
22- 18 23 23 - 21 = 2 18 × 2 = 36
24- 13 25 25 - 21 = 4 13 × 4 = 52
26- 3 27 27 - 21 = 6 3 × 6 = 18
28- 30 2 29 29 - 21 = 8 2 × 8 = 16
X 0
= X-X0
E
= X-X0
E
∑ f. E
∑ f
X = X 0
+ ‫ـــــــــــــ‬
82
100
X = 21 + ‫ـــــــــ‬ = 21 + 0.82
X =21 .82
.‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﻧﺴﺘﺨﺮج‬ (
. ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫أﻛﺒﺮ‬ ‫ﻳﻘﺎﺑﻞ‬ ‫اﻟﺬي‬ (21) ‫وﻟﻴﻜﻦ‬ ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ ( ) ‫اﻟﻔﺮﺿﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻧﺨﺘﺎر‬ (X) ‫وﻟﻴﻜﻦ‬ ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ (X) ‫وﻟﻴﻜﻦ‬ ‫اﻟﻔﺌﺎت‬ ‫اﻛﺰ‬‫ﺮ‬‫ﻣ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ (
‫اﻻﻋﻤﺎر‬ 18 20 22 24 26 28-30 ‫اﻟﻤﺠﻤﻮع‬
‫ﻋﺪد‬
‫اﻟﻄﻼب‬
20 44 18 13 3 2 100

142
: ‫وعيوبه‬ ‫الحسابي‬ ‫الوسﻂ‬ ‫ايا‬‫ز‬‫م‬
: ‫اﻳــﺎ‬‫ﺰ‬‫اﻟﻤ‬
. ‫اﻟﺒﺴﻴﻄﺔ‬ ‫اﻟﺤﺴﺎﺑﻴﺔ‬ ‫ﺑﻌﻤﻠﻴﺎﺗﻪ‬ ‫ﻳﺘﻤﻴﺰ‬ (1)
. ‫ﺣﺴﺎﺑﻪ‬ ‫ﻓﻲ‬ ‫اﻟﻘﻴﻢ‬ ‫ﺟﻤﻴﻊ‬ ‫ﺗﺪﺧﻞ‬ (2)
: ‫اﻟﻌﻴﻮب‬
. ً‫ا‬‫ﺟﺪ‬ ‫اﻟﺼﻐﻴﺮة‬ ‫او‬ ً‫ا‬‫ﺟﺪ‬ ‫اﻟﻜﺒﻴﺮة‬ ‫اﻟﻤﺘﻄﺮﻓﺔ‬ ‫او‬ ‫اﻟﺸﺎذة‬ ‫ﺑﺎﻟﻘﻴﻢ‬ ‫ﻳﺘﺄﺛﺮ‬ (1)
. ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ً‫ﺎ‬‫ﺣﺴﺎﺑ‬ ‫ﺣﺴﺎﺑﻪ‬ ‫ﻳﻤﻜﻦ‬ ‫ﻻ‬ (2)
� Median ‫اﻟﻮﺳﻴﻂ‬ [7 � 3�
(7-2) ‫ﺗﻌﺮﻳﻒ‬
‫أو‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﻳ‬ ‫ﺗﺮﺗﻴﺒﻬﺎ‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺗﺘﻮﺳﻂ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺑﺄﻧﻪ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻮﺳﻴﻂ‬ ‫ﻳﻌﺮف‬
. ‫ﻣﻨﻪ‬ ‫اﻻﻛﺒﺮ‬ ‫ﻟﻠﻘﻴﻢ‬ ً‫ﺎ‬‫ﻣﺴﺎوﻳ‬ ‫ﻳﻜﻮن‬ ‫ﻣﻨﻪ‬ ‫اﻷﺻﻐﺮ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻋﺪد‬ ‫ﻓﺎن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ً‫ﺎ‬‫ﺗﻨﺎزﻟﻴ‬
: ‫الوسيﻂ‬ ‫حساب‬ ‫ﻃريقة‬
: ‫المبوبة‬ ‫ﻏير‬ ‫البيانات‬ (١
.‫اﻟﻮﺳﻴﻂ‬ ‫ﻫﻲ‬ ‫ﻟﺘﻜﻮن‬ ‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻓﻲ‬ ‫ﺗﻘﻊ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻧﺄﺧﺬ‬ ‫ﺛﻢ‬ ً‫ﺎ‬‫ﺗﻨﺎزﻟﻴ‬ ‫أو‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﻳ‬ ً‫ﺎ‬‫ﺗﺮﺗﻴﺒ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻧﺮﺗﺐ‬
. ‫ﻓﺮدي‬ ‫اﻟﻘﻴﻢ‬ ‫ﻋﺪد‬ ‫أن‬ ‫ﺑﻔﺮض‬ ‫ﻫﺬا‬
‫ﻣﺠﻤﻮع‬ ‫ﻫﻮ‬ ‫اﻟﻮﺳﻴﻂ‬ ‫وﻳﻜﻮن‬ ‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻓﻲ‬ ‫اﻟﻠﺘﻴﻦ‬ ‫اﻟﻘﻴﻤﺘﻴﻦ‬ ‫ﻓﻨﺄﺧﺬ‬ ً‫ﺎ‬‫زوﺟﻴ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻋﺪد‬ ‫ﻛﺎن‬ ‫إذا‬ ‫أﻣﺎ‬
.‫اﺛﻨﻴﻦ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫ﻣﻘﺴﻮﻣ‬ ‫اﻟﻘﻴﻤﺘﻴﻦ‬

143
� � 6 ‫ﻣﺜﺎل‬
.‫ﻛﻐﻢ‬ 55 ،‫ﻛﻐﻢ‬ 63 ،‫05ﻛﻐﻢ‬ ،‫85ﻛﻐﻢ‬ ،‫ﻛﻐﻢ‬ 52 : ‫ﻫﻲ‬ ‫واﻟﺘﻲ‬ ‫اﻟﻄﻼب‬ ‫ﺑﻌﺾ‬ ‫ان‬‫ز‬‫ﻷو‬ ‫اﻟﻮﺳﻴﻂ‬ ‫اﺣﺴﺐ‬
‫ﻓﻲ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫ﻫﻲ‬ ‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻓﻲ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫أن‬ ‫ﻧﻼﺣﻆ‬ :‫ﺗﺼﺎﻋﺪﻳﺎ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻧﺮﺗﺐ‬
‫اﻟﺘﺮﺗﻴﺐ‬
55= ‫اﻟﻮﺳﻴﻂ‬ ∴
� � 7 ‫ﻣﺜﺎل‬
،‫ﻛﻐﻢ‬ 57 ،‫ﻛﻐﻢ‬ 63 ،‫05ﻛﻐﻢ‬ ،‫85ﻛﻐﻢ‬ ،‫ﻛﻐﻢ‬ 52 :‫اﻟﻄﻼب‬ ‫ﻟﺒﻌﺾ‬ ‫اﻟﺘﺎﻟﻴﺔ‬ ‫ان‬‫ز‬‫ﻟﻸو‬ ‫اﻟﻮﺳﻴﻂ‬ ‫اﺣﺴﺐ‬
.‫55ﻛﻐﻢ‬
:ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﻳ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻧﺮﺗﺐ‬
‫وﻳﻜﻮن‬ ‫اﻟﻤﻨﺘﺼﻒ‬ ‫ﻓﻲ‬ ‫ﻗﻴﻤﺘﻴﻦ‬ ‫وﺟﻮد‬ ‫ﻧﻼﺣﻆ‬
(‫)اﻟﺜﺎﻟﺚ‬ 3 = ‫ــــــــــــــ‬ = ‫ـــــــــــــــــــ‬ =‫اﻻول‬ ‫ﺗﺮﺗﻴﺐ‬
(‫اﺑﻊ‬‫ﺮ‬‫)اﻟ‬ 4 = 1 + 3 = 1 + ‫ـــــــــــــــ‬ =‫اﻟﺜﺎﻧﻲ‬ ‫ﺗﺮﺗﻴﺐ‬
56 = ‫ــــــــــــــــــــــــــــ‬ = ‫ــــــــــــــــــــــــــــــــــــــ‬ =‫اﻟﻮﺳﻴﻂ‬ ∴
50، 52، 55، ، 63
50، 52، ، 58، 63
6
2
n
2
n
2
‫اﺑﻊ‬‫ﺮ‬‫اﻟ‬ + ‫اﻟﺜﺎﻟﺚ‬
2
57 + 55
2
58
55 ، 57

144
‫اﻟﻮزن‬ ‫ﻓﺌﺎت‬‫اﻻﺷﺨﺎص‬ ‫ﻋﺪد‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺠﺘﻤﻊ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬
30-99
40-1524
50-2246
60-2571
70-1889
80 - 9011100
‫اﳌﺠﻤﻮع‬100
+
+
+
+
+
‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫ﻗﺒﻞ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺘﺠﻤﻊ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ - ‫اﻟﻮﺳﻴﻂ‬ ‫ﺗﺮﺗﻴﺐ‬
‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﻣﺠﻤﻮﻋﺔ‬
2
‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬
: ‫المبوبة‬ ‫البيانات‬ ‫في‬ �2
: ‫ﻳﺄﺗﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﺤﻞ‬ ‫ﺧﻄﻮات‬ ‫وﺗﻜﻮن‬ :‫اﻟﻔﺌﺎت‬ ‫ذات‬ ‫ﺑﺔ‬ّ‫ﻮ‬‫اﻟﻤﺒ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻮﺳﻴﻂ‬ ‫ﺣﺴﺎب‬ ‫ﻳﻤﻜﻦ‬
.‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺘﺠﻤﻊ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﺟﺪول‬ ‫ﻧﻜﻮن‬ (1
‫ــــــــــــــــــــــــــــ‬ = ‫اﻟﻮﺳﻴﻂ‬ ‫ﺗﺮﺗﻴﺐ‬ ‫ﺣﺴﺎب‬ (2
‫اﻟﻔﺌﺔ‬ ‫وﺗﺴﻤﻰ‬ ‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺘﺠﻤﻊ‬ ‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﻮﺳﻴﻂ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺘﻮي‬ ‫اﻟﺘﻲ‬ ‫اﻟﻔﺌﺔ‬ ‫ﺗﺤﺪﻳﺪ‬ (3
. ‫اﻟﻮﺳﻴﻂ‬ ‫ﺗﺮﺗﻴﺐ‬ ‫ﻳﺴﺎوي‬ ‫أو‬ ‫أﻛﺒﺮ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫أول‬ ‫ﺗﻘﺎﺑﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻔﺌﺔ‬ ‫وﻫﻲ‬ ‫اﻟﻮﺳﻄﻴﺔ‬
‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫ﻗﺒﻞ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺠﺘﻤﻊ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ -‫اﻟﻮﺳﻴﻂ‬ ‫ﺗﺮﺗﻴﺐ‬
‫اﻟﻔﺌﺔ‬ ‫ﻃﻮل‬ × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ + ‫اﻟﻮﺳﻴﻄﻴﻪ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻷدﻧﻰ‬ ‫اﻟﺤﺪ‬ = ‫اﻟﻮﺳﻴﻂ‬
‫اﻟﺼﺎﻋﺪ‬ ‫اﻟﻤﺘﺠﻤﻊ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ fb ، ME = ‫اﻟﻮﺳﻴﻂ‬ ‫ﺣﻴﺚ‬ ME = L + ‫ـــــــــــــــــــــــ‬ . W
‫ﻟﻠﻠﻔﺌﺔ‬ ‫اﻻدﻧﻰ‬ ‫اﻟﺤﺪ‬ :L ، ‫اﻟﻔﺌﺔ‬ ‫:ﻃﻮل‬W ، ‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ : fm ، ‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫ﻗﺒﻞ‬ ‫ﻟﻠﻔﺌﺔ‬
. ‫اﻟﻮﺳﻴﻄﻴﺔ‬
� � 8 ‫ﻣﺜﺎل‬
‫ﻣﻦ‬ ‫اﻟﻮزن‬ ‫وﺳﻴﻂ‬ ‫ﺟﺪ‬
: ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬
50 = ‫ــــــــــ‬ = ‫اﻟﻮﺳﻴﻂ‬ ‫ﺗﺮﺗﻴﺐ‬
( 60 - 70 ) = ‫اﻟﻮﺳﻴﻄﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ∴
ME = L + ‫ـــــــــــــــــــــــ‬ . W
ME = 60 + ‫ــــــــــــــــــ‬ × 10
= 60 + ‫ـــــــــ‬ ME = 60 + 1.6 = 61.6
∑ f‫ــــــ‬2 - f b
f m
100
2
∑ f‫ــــــ‬2 - f b
f m
50 46-
25
8 ⇐5

145
: ‫وعيوبه‬ ‫الوسيﻂ‬ ‫ايا‬‫ز‬‫م‬
‫اﻟﻤﺘﻄﺮﻓﺔ‬ ‫أو‬ ‫اﻟﺸﺎذة‬ ‫ﺑﺎﻟﻘﻴﻢ‬ ‫ﻻﻳﺘﺄﺛﺮ‬ (1)
.ً‫ﺎ‬‫ﺑﻴﺎﻧﻴ‬ ً‫ﺎ‬‫ﺣﺴﺎﺑ‬ ‫ﺣﺴﺎﺑﻪ‬ ‫ﻳﻤﻜﻦ‬ (2)
.‫ﺣﺴﺎﺑﻪ‬ ‫ﻓﻲ‬ ‫اﻟﻘﻴﻢ‬ ‫ﺟﻤﻴﻊ‬ ‫ﻻﺗﺪﺧﻞ‬ (1)
. ‫اﻟﺘﻘﺮﻳﺒﻴﺔ‬ ‫ﺑﺎﻟﻄﺮق‬ ‫ﺣﺴﺎﺑﻪ‬ ‫ﻳﻜﻮن‬ ‫اﻟﻔﺌﺎت‬ ‫ذات‬ ‫اﻟﻤﺒﻮﺑﺔ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ (2)
� Mode ‫اﻟﻤﻨﻮال‬ [7�4�
(7 - 3)‫ﺗﻌﺮﻳﻒ‬
‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ .‫ات‬‫ر‬‫ا‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫أﻛﺒﺮ‬ ‫ﺗﻘﺎﺑﻞ‬ ‫اﻟﺘﻲ‬ ‫أو‬ ً‫ا‬‫ر‬‫ا‬‫ﺮ‬‫ﺗﻜ‬ ‫اﻷﻛﺜﺮ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﺑﺄﻧﻪ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻟﻤﺠﻤﻮﻋﺔﻣﻦ‬ ‫اﻟﻤﻨﻮال‬ ‫ف‬ّ‫ﺮ‬َ‫ﻌ‬ُ‫ﻳ‬
� ‫اﻟﻤﻨﻮال‬ ‫ﺣﺴﺎب‬ ‫ﻃﺮﻳﻘﺔ‬
� ‫ﺑﺔ‬ ّ‫اﻟﺒﻮ‬ ‫ﻏﻴﺮ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ (1
� � 9 ‫ﻣﺜﺎل‬
: ‫اﻵﺗﻴﺔ‬ ‫اﻷﻋﺪاد‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻫﻲ‬ ‫ﻣﺎ‬
4 ،2 ،4 ،7 ،8 ،3 ،4 ،9 ،7 ،4 ( ‫أ‬
. ‫ﻏﻴﺮﻫﺎ‬ ‫ﻣﻦ‬ ‫أﻛﺜﺮ‬ ‫ﺗﻜﺮرت‬ ‫ﻻﻧﻬﺎ‬ 4 = ‫اﻟﻤﻨﻮال‬
18 ،10 ،5 ،6 ،8 ،1 ،5 ،6 (‫ب‬
.‫ﻏﻴﺮﻫﻤﺎ‬ ‫ﻣﻦ‬ ‫أﻛﺜﺮ‬ ‫ا‬‫ر‬‫ﺗﻜﺮ‬ ‫ﻷﻧﻬﻤﺎ‬ 6 ،5 = ‫اﻟﻤﻨﻮال‬
12 ،11 ،10 ،7 ،3 ،4 ،5 ،8 (‫ﺟـ‬
‫ﻻﻳﻮﺟﺪ‬ = ‫اﻟﻤﻨﻮال‬
MO

146
: ‫المبوبة‬ ‫البيانات‬ ( 2
: (‫ﺑﻴﺮﺳﻮن‬ ‫)ﻃﺮﻳﻘﺔ‬ ‫اﻟﻔﺮوق‬ ‫ﻃﺮﻳﻘﺔ‬ ( ‫أ‬
‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ﻃﻮل‬ × ‫ـــــــــــــــ‬ + ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻻدﻧﻰ‬ ‫اﻟﺤﺪ‬ = ‫اﻟﻤﻨﻮال‬
.‫ﻗﺒﻠﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ - ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ = d1 ‫ﺣﻴﺚ‬
.‫ﺑﻌﺪﻫﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ - ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ =d2
.‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫أﻛﺒﺮ‬ ‫ﺗﻘﺎﺑﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫واﻟﻔﺌﺔ‬ .‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺠﺪول‬ ‫ﻓﻲ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫أﻛﺒﺮ‬ ‫ﻫﻮ‬ ‫اﻟﻤﻨﻮاﻟﻲ‬ ‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫وإن‬
� � 10 ‫ﻣﺜﺎل‬
: ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﻤﻨﻮال‬ ‫اﺣﺴﺐ‬
d1 = 25 - 22 =3
d2 = 25 - 18 = 7
10 = 70 - 60 = ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ﻃﻮل‬
‫اﻟﻔﺌﺔ‬ ‫×ﻃﻮل‬ ‫ـــــــــــ‬ + ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻻدﻧﻰ‬ ‫اﻟﺤﺪ‬ = ‫اﻟﻤﻨﻮال‬
10 × ‫ــــــــــــــــ‬ + 60 = ‫اﻟﻤﻨﻮال‬
3 + 60 = ‫اﻟﻤﻨﻮال‬
63 = ‫اﻟﻤﻨﻮال‬
: (‫اﻓﻌﺔ‬‫ﺮ‬‫)اﻟ‬ ‫اﻟﻌﺰوم‬ ‫ﻃﺮﻳﻘﺔ‬ (‫ب‬
.‫اﻟﻌﺘﻠﺔ‬ ‫ﻧﻬﺎﻳﺘﻲ‬ ‫إﺣﺪى‬ ‫ﻋﻨﺪ‬ ‫ﺗﺆﺛﺮ‬ ‫ﻗﻮة‬ ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫وﻧﺠﻌﻞ‬ ‫ﻋﺘﻠﺔ‬ ‫ﻧﺮﺳﻢ‬ ‫اﻟﻄﺮﻳﻘﺔ‬ ‫ﻫﺬه‬ ‫ﻓﻲ‬ (1
‫ﻃﻮل‬ = ‫اﻟﻌﺘﻠﺔ‬ ‫وﻃﻮل‬ ‫ﻟﻠﻌﺘﻠﺔ‬ ‫اﻷﺧﺮى‬ ‫اﻟﻨﻬﺎﻳﺔ‬ ‫ﺗﺆﺛﺮﻋﻨﺪ‬ ‫ﻗﻮة‬ ‫اﻟﻤﻨﻮاﻟﻴﻪ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﻟﺘﻜ‬ ‫اﻟﻼﺣﻖ‬ ‫ار‬‫ﺮ‬‫واﻟﺘﻜ‬
‫اﻟﻔﺌﺔ‬
x =‫اﻟﻄﺮﻓﻴﻦ‬ ‫أﺣﺪ‬ ‫ﻋﻨﺪ‬ ‫اﻟﻤﻨﻮال‬ ‫ﻌﺪ‬ُ‫ﺑ‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺘﻲ‬ ‫اﻻرﺗﻜﺎز‬ ‫ﻧﻘﻄﺔ‬ ‫ﻧﻔﺮض‬ (2
.(‫اﻋﻬﺎ‬‫ر‬‫ذ‬ × ‫اﻟﻤﻘﺎوﻣﺔ‬ = ‫اﻋﻬﺎ‬‫ر‬‫×ذ‬ ‫اﻟﻘﻮة‬ ) ‫اﻟﻌﺘﻠﺔ‬ ‫ﻗﺎﻧﻮن‬ ‫ﻧﻄﺒﻖ‬ (3
. ‫اﻟﻤﻨﻮال‬ ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻷدﻧﻰ‬ ‫اﻟﺤﺪ‬ ‫إﻟﻰ‬ ‫وﻧﻀﻴﻔﻬﺎ‬ x ‫ﻗﻴﻤﺔ‬ ‫ﻧﺴﺘﺨﺮج‬ (4
‫اﻟﻔﺌﺔ‬ ‫×ﻃﻮل‬ ‫ـــــــــــ‬ + ‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻻدﻧﻰ‬ ‫اﻟﺤﺪ‬ = ‫اﻟﻤﻨﻮال‬
‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫ﻓﺌﺎت‬
9 30-
15 40-
22 50-
25 60-
18 70-
11 80-90
‫اﻟﻼﺣﻖ‬ ‫اﻟﺘﻜﺮار‬
‫اﳌﻨﻮاﻟﻲ‬ ‫اﻟﺘﻜﺮار‬
‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺘﻜﺮار‬
d1
d1 +d2
d1
d1+d2
3 + 7
3

147
� � 11 ‫ﻣﺜﺎل‬
: ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﻤﻨﻮال‬ ‫ﺟﺪ‬
(70-60) =‫اﻟﻤﻨﻮاﻟﻴﺔ‬ ‫اﻟﻔﺌﻪ‬
10 = ‫اﻟﻔﺌﺔ‬ ‫ﻃﻮل‬ =‫اﻟﻌﺘﻠﺔ‬ ‫ﻃﻮل‬
‫اﻋﻬﺎ‬‫ر‬‫ذ‬ × ‫اﻟﻤﻘﺎوﻣﺔ‬ = ‫اﻋﻬﺎ‬‫ر‬‫×ذ‬ ‫اﻟﻘﻮة‬ ∵
(10 -x) (37)= x (38)
370 - 37x = 38x
75x = 370
x = ‫ـــــــــــــــ‬ = 4.9 ∴
64.9 = 4.9 + 60 = ‫اﻟﻤﻨﻮال‬ ∴
: ‫وعيوبه‬ ‫المنوال‬ ‫ايا‬‫ز‬‫م‬
‫ﺣﺴﺎﺑﻪ‬ ‫ﻃﺮﻳﻘﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺴﻴﻂ‬ (1)
.‫واﻟﻤﺘﻄﺮﻓﺔ‬ ‫اﻟﺸﺎذة‬ ‫ﺑﺎﻟﻘﻴﻢ‬ ‫ﻻﻳﺘﺄﺛﺮ‬ (2)
‫اﻟﺘﻘﺮﻳﺒﻴﺔ‬ ‫ﺑﺎﻟﻄﺮق‬ ‫ﺣﺴﺎﺑﻪ‬ ‫ﻳﻜﻮن‬ ‫اﻟﻔﺌﺎت‬ ‫ذات‬ ‫اﻟﻤﺒﻮﺑﺔ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ (1)
.‫ﻏﻴﺮﻫﺎ‬ ‫ﻣﻦ‬ ‫اﻛﺜﺮ‬ ‫ﻣﺘﻜﺮرة‬ ‫ﻗﻴﻢ‬ ‫وﺟﻮد‬ ‫ﻋﺪم‬ ‫ﺣﺎﻟﻪ‬ ‫ﻓﻲ‬ ‫اﻳﺠﺎده‬ ‫ﻻﻳﻤﻜﻦ‬ (2)
‫اﻟﺪرﺟﺔ‬ ‫ﺑﻨﻔﺲ‬ ‫اﻟﻘﻴﻢ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻣﻨﻮال‬ ‫ﻣﻦ‬ ‫اﻛﺜﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻗﺪ‬ (3)
90 -10080-70-60-50-40-‫اﻟﻔﺌﺎت‬
283759386‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬
‫ﺑﻌﺪ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬
37 = ‫اﻟﻤﻨﻮاﻟﻴﺔ‬
‫ﻗﺒﻞ‬ ‫اﻟﻔﺌﺔ‬ ‫ار‬‫ﺮ‬‫ﺗﻜ‬
38 = ‫اﻟﻤﻨﻮاﻟﻴﺔ‬
10 = ‫اﻟﻔﺌﺔ‬ ‫ﻃﻮل‬ = ‫اﻟﻌﺘﻠﺔ‬ ‫ﻃﻮل‬
10 -x x
‫اﻻرﺗﻜﺎز‬
370
75

148
( 7 - 1 ) ‫تمرينات‬
. ‫واﻟﻤﻨﻮال‬ ‫واﻟﻮﺳﻴﻂ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻋﺮف‬ / 1‫س‬
،17 ،18 ،17 ،15 ،16 ،18 ،16 ،17 ،15 . ‫اﻟﻄﻼب‬ ‫ﻣﻦ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫أﻋﻤﺎر‬ ‫ﺗﻤﺜﻞ‬ ‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ / 2‫س‬
‫اﻟﻤﻨﻮال‬ ( ‫ﺟـ‬ ، ‫اﻟﻮﺳﻴﻂ‬ (‫ب‬ ، ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ( ‫أ‬ :‫ﺟﺪ‬ 19
‫ﻣﺠﻤﻮع‬ ‫دﻳﻨﺎر.ﻓﻤﺎ‬ (40000) ‫أﺷﺨﺎص‬ ‫ﻟﺨﻤﺴﺔ‬ ‫اﻟﺸﻬﺮي‬ ‫ﻟﻠﺪﺧﻞ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻛﺎن‬ ‫إذا‬ / 3‫س‬
‫؟‬ ‫دﺧﻮﻟﻬﻢ‬
‫ﻓﺼﻞ‬ ‫ﻓﻲ‬ ً‫ﺎ‬‫ﻳﻮﻣ‬ 90 ‫ﺧﻼل‬ ‫اﻟﻤﺪن‬ ‫إﺣﺪى‬ ‫ﻓﻲ‬ ‫ارة‬‫ﺮ‬‫اﻟﺤ‬ ‫درﺟﺎت‬ ‫ﻣﺠﻤﻮع‬ ‫ﻳﺒﻦ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ / 4‫س‬
: ‫اﻷﻋﻮام‬ ‫أﺣﺪ‬ ‫ﻓﻲ‬ ‫اﻟﺼﻴﻒ‬
.‫اﻟﻤﻨﻮال‬ (‫ﺟـ‬ .‫اﻟﻮﺳﻴﻂ‬ (‫ب‬ .‫ارة‬‫ﺮ‬‫اﻟﺤ‬ ‫ﻟﺪرﺟﺎت‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ( ‫أ‬ : ‫اﻟﻤﻄﻠﻮب‬
:‫اﻟﺮواﺗﺐ‬ ‫ﻟﻬﺬه‬ ‫اﻟﻮﺳﻴﻂ‬ ‫إﻳﺠﺎد‬ ‫واﻟﻤﻄﻠﻮب‬ ‫ﻣﺪرﺳﺔ‬ ‫ﻓﻲ‬ ً‫ﺎ‬‫ﻣﻌﻠﻤ‬ 60 ‫رواﺗﺐ‬ ‫ﻳﺒﻴﻦ‬ ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬ / 5‫س‬
‫اﻟﻮﺳﻂ‬ ‫ﺟﺪ‬ ‫اﻟﻤﺪن‬ ‫إﺣﺪى‬ ‫ﻓﻲ‬ ‫اﻟﻤﺤﻼت‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮع‬ ‫اﻟﻴﻮﻣﻴﺔ‬ ‫اﻷرﺑﺎح‬ ‫ﻳﺒﻴﻦ‬ ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬ / 6‫س‬
: ‫اﻻرﺑﺎح‬ ‫ﻟﻬﺬه‬ (‫اﻟﻴﻮﻣﻲ‬ ‫اﻟﺮﺑﺢ‬ ‫)ﻣﻌﺪل‬ ‫اﻟﺤﺴﺎﺑﻲ‬
‫اﻟﻤﺠﻤﻮع‬44-4840-36-32-28-24-20-‫ارة‬‫ﺮ‬‫اﻟﺤ‬ ‫درﺟﺎت‬ ‫ﻓﺌﺎت‬
9079152318108‫اﻻﻳﺎم‬ ‫ﻋﺪد‬
200-210190-180-170-160-150-‫دﻳﻨﺎر‬ ‫ﺑﺎﻟﻒ‬ ‫اﺗﺐ‬‫ﺮ‬‫اﻟ‬
372015105‫اﳌﻌﻠﻤﻦﻴ‬ ‫ﻋﺪد‬
24-2820-16-12-8-4-‫دﻳﻨﺎر‬ ‫ﺑﺎﻟﻒ‬ ‫اﻟﻴﻮﻣﻲ‬ ‫اﻟﺮﺑﺢ‬
6122015108‫اﳌﺤﻼت‬ ‫ﻋﺪد‬

149
� Measusres of Variation ‫اﻟﺘﺸﺘﺖ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ [7�5�
‫ﻣﺘﺠﻤﻌﺔ‬ ‫ﺗﻜﻮن‬ ‫رﺑﻤﺎ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻫﺬه‬ ‫أﻋﺪاد‬ ‫وإن‬ ،‫ﺣﺴﺎﺑﻴﺎ‬ ً‫ﺎ‬‫وﺳﻄ‬ ‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻜﻞ‬ ‫إن‬
‫ﻓﺎن‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫وﺳﻄﻬﺎ‬ ‫ﻣﻦ‬ ‫ﺑﺎﻟﻘﺮب‬ ‫ﻣﺘﺠﻤﻌﺔ‬ ‫اﻻﻋﺪاد‬ ‫ﻫﺬه‬ ‫ﻛﺎﻧﺖ‬ ‫ﻓﺎذا‬ .‫ﻋﻨﻪ‬ ‫ﻣﺒﺘﻌﺪة‬ ‫أو‬ ‫ﻣﻨﻪ‬ ‫ﺑﺎﻟﻘﺮب‬
‫ﻛﺒﻴﺮ‬ ‫ﺗﺸﺘﺘﻬﺎ‬ ‫ﻓﺎن‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫وﺳﻄﻬﺎ‬ ‫ﻋﻦ‬ ‫ﻣﺒﺘﻌﺪة‬ ‫اﻷﻋﺪاد‬ ‫ﻫﺬه‬ ‫ﻛﺎﻧﺖ‬ ‫وإذا‬ ،‫ﺿﺌﻴﻞ‬ ‫ﺗﺸﺘﺘﻬﺎ‬ ‫ﻣﻘﺪار‬
50 ‫ﻫﻮ‬ 70 ،60 ،50 ،40 ،30 ‫ﻟﻸﻋﺪاد‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫أن‬ :ً‫ﻼ‬‫ﻣﺜ‬
50 ‫ﻫﻮ‬ 30 ،100 ،90 ،20 ،10 :‫ﻟﻸﻋﺪاد‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫واﻟﻮﺳﻂ‬
‫أﻋﺪاد‬ ‫ﺗﺘﺸﺘﺖ‬ ‫ﺑﻴﻨﻤﺎ‬ ‫ﺿﺌﻴﻞ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻋﻦ‬ ‫ﺗﺸﺘﺘﻬﺎ‬ ‫أن‬ ‫ﺗﺸﺎﻫﺪ‬ ‫اﻷوﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺗﺄﻣﻞ‬ ‫ﻋﻨﺪ‬
. ‫ﻛﺒﻴﺮ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻋﻦ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬
: ‫التﺸتﺖ‬ ‫مقاييس‬
: ‫ﻫﻲ‬ ‫ﻧﺪرﺳﻬﺎ‬ ‫ﺳﻮف‬ ‫اﻟﺘﻲ‬ ‫اﻟﺘﺸﺘﺖ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ ‫إن‬
. Range ‫اﻟﻤﺪى‬ - 1
. Standard Deviation ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ - 2
‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫ﻗﻴﻤﺔ‬ ‫وأﺻﻐﺮ‬ ‫ﻗﻴﻤﺔ‬ ‫أﻛﺒﺮ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻔﺮق‬ ‫ﻫﻮ‬ �‫اﻟﻤﺪى‬ [7�5�1�
‫وﻫﻤﺎ‬ . ‫اﻟﻤﺘﻐﻴﺮ‬ ‫ﻗﻴﻢ‬ ‫ﻣﻦ‬ ‫ﻓﻘﻂ‬ ‫ﻗﻴﻤﺘﻴﻦ‬ ‫ﻋﻠﻰ‬ ‫ﻳﺘﻮﻗﻒ‬ ‫ﻻﻧﻪ‬ ‫ﻟﻠﺘﺸﺘﺖ‬ ‫ﻣﻬﻢ‬ ‫ﻣﻘﻴﺎس‬ ‫ذات‬ ‫ﻟﻴﺲ‬ ‫واﻟﻤﺪى‬
‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫ﻳﺤﺪث‬ ‫ﺗﻐﻴﺮ‬ ‫أي‬ ‫وإن‬ ‫اﻟﻌﻴﻨﺔ‬ ‫ﺑﺬﺑﺬﺑﺎت‬ ً‫ﺎ‬‫ﺑﺎﻟﻐ‬ً‫ا‬‫ﺮ‬‫ﺗﺄﺛ‬ ‫ﻳﺘﺄﺛﺮ‬ ‫ﻓﻬﻮ‬ ‫وﻟﺬا‬ ،‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫ﻗﻴﻤﺔ‬ ‫وأﻛﺒﺮ‬ ‫أﻗﻞ‬
. ‫اﻟﻤﺪى‬ ‫ﻗﻴﻤﺔ‬ ‫ﻓﻲ‬ ‫ﺑﻮﺿﻮح‬ ‫ﻳﺆﺛﺮ‬ ‫اﻟﻘﻴﻤﺘﻴﻦ‬ ‫ﻫﺎﺗﻴﻦ‬
: ‫ﺑﺔ‬ّ‫ﻮ‬‫ﻣﺒ‬ ‫ﻏﻴﺮ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ (‫أ‬
� � 12 ‫ﻣﺜﺎل‬
98 ،24 ،68 ،35 ،12 :‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﻘﻴﻢ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺪى‬ ‫ﻣﺎﻫﻮ‬
R = 98 - 12 = 86 ‫اﳌﺪى‬

150
:‫ﺑﺔ‬ّ‫ﻮ‬‫ﻣﺒ‬ ‫اﻟﺒﻴﺎﻧﺎت‬ (‫ب‬
� � 13 ‫ﻣﺜﺎل‬
: ‫اﻟﺘﺎﻟﻲ‬ ‫اري‬‫ﺮ‬‫اﻟﺘﻜ‬ ‫اﻟﺘﻮزﻳﻊ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺪى‬ ‫ﻣﺎﻫﻮ‬
55-5 ‫اﻷوﻟﻰ‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻷدﻧﻰ‬ ‫اﻟﺤﺪ‬ - ‫اﻻﺧﻴﺮة‬ ‫ﻟﻠﻔﺌﺔ‬ ‫اﻷﻋﻠﻰ‬ ‫اﻟﺤﺪ‬ = ‫اﻟﻤﺪى‬
∴ R = 50
� ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ [7�5�2�
‫اﻟﻤﻔﺮدات‬ ‫ﻣﻦ‬ n ‫ﻟﺪﻳﻨﺎ‬ ‫ﻛﺎﻧﺖ‬ ‫ﻓﺎذا‬ .ً‫ﺎ‬‫إﺳﺘﺨﺪاﻣ‬ ‫اﻟﺘﺸﺘﺖ‬ ‫ﻣﻘﺎﻳﻴﺲ‬ ‫أﻛﺜﺮ‬ ‫ﻣﻦ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫ﻳﻌﺪ‬
‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫ﺑﻌﻀﻬﺎ‬ ‫ﻣﻦ‬ ‫ﻣﺘﻘﺎرﺑﺔ‬ ‫ﺗﻜﻮن‬ ‫اﻟﻤﻔﺮدات‬ ‫ﻫﺬه‬ ‫ﻓﺎن‬ . x ‫اﻟﺤﺴﺎﺑﻲ‬ ‫ووﺳﻄﻬﺎ‬ x1
، x2
، ....، xn
‫ﻓﺎن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ، x1
، x2
,.... ,‫ﺻﻐﻴﺮة‬ x ‫ﻋﻦ‬ ‫اﻓﺎﺗﻬﺎ‬‫ﺮ‬‫إﻧﺤ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫أي‬ x ‫اﻟﺤﺴﺎﺑﻲ‬ ‫وﺳﻄﻬﺎ‬ ‫ﻣﻦ‬ ‫ﻗﺮﻳﺒﺔ‬
‫ذﻟﻚ‬ ‫ﻳﺘﻢ‬ ‫أن‬ ‫وﻳﻤﻜﻦ‬ .‫اﻟﺘﺸﺘﺖ‬ ‫ﻟﻘﻴﺎس‬ ‫اﺳﺘﺨﺪاﻣﻬﺎ‬ ‫ﻳﻤﻜﻦ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫وﺳﻄﻬﺎ‬ ‫ﻋﻦ‬ ‫اﻟﻤﻔﺮدات‬ ‫اﻓﺎت‬‫ﺮ‬‫إﻧﺤ‬
.‫اﻓﺎت‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫ﻫﺬه‬ ‫ﻣﺘﻮﺳﻂ‬ ‫ﺑﺄﺧﺬ‬
(7-4)‫ﺗﻌﺮﻳﻒ‬
‫ﻣﻔﺮدات‬ ‫ﻗﻴﻢ‬ ‫اﻓﺎت‬‫ﺮ‬‫إﻧﺤ‬ ‫ﻣﺮﺑﻌﺎت‬ ‫ﻟﻤﺘﻮﺳﻂ‬ ‫اﻟﺘﺮﺑﻴﻌﻲ‬ ‫ﻟﻠﺠﺬر‬ ‫اﻟﻤﻮﺟﺒﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻫﻮ‬ : ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬
. (S) ‫ﺑﺎﻟﺮﻣﺰ‬ ‫ﻟﻪ‬ ‫وﻳﺮﻣﺰ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫وﺳﻄﻬﺎ‬ ‫ﻋﻦ‬ ‫اﻟﺘﻮزﻳﻊ‬
: ‫ﻣﺒﻮﺑﺔ‬ ‫ﻏﻴﺮ‬ ‫ﻟﻘﻴﻢ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫ﺣﺴﺎب‬
S = - ( x )2
45-5535-25-15-5-‫اﻟﻔﺌﺎت‬
7141583‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬
∑ x2
‫ــــــــ‬n√

151
‫ﻣﻦ‬ ‫ﺛـﺎﺑﺘﺔ‬ ‫ﻛﻤﻴـــﺔ‬ ‫ﻃـــﺮح‬ ‫ﻋﻨﺪ‬
‫ﻗﻴﻤﺔ‬‫ﻋﻠﻰ‬‫ﻻﺗﺆﺛﺮ‬،‫اﻟﻘﻴﻢ‬‫ﺟﻤﻴﻊ‬
‫واﻟﻤﺜـــﺎل‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻹﻧﺤ‬
: ‫ذﻟﻚ‬ ‫ﻳﻮﺿﺢ‬ (15)
∑ x2
25
5
x1
+ x2
...+ xn
n
√165
5 √
√ √
20
55
8 + 6 + 4 + 2 + 0
∑ x2
√120
5 √
√ √
� � 14 ‫ﻣﺜﺎل‬
9 ،7 ،5 ،3 ،1 :‫اﻵﺗﻴﺔ‬ ‫ﻟﻠﻘﻴﻢ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫إﺣﺴﺐ‬
X = ‫ــــــــــــــــــــــــ‬ = ‫ــــــــــــــ‬ = 5
S = - ( x )2
S = ‫ــــــــ‬ - 25 = 33 - 25
‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ .... 8 = 2 2
� � 15 ‫ﻣﺜﺎل‬
‫اﻟﻨﺘﻴﺠﺔ‬ ‫ﻗﺎرن‬ . ‫اﻟﺠﺪﻳﺪة‬ ‫ﻟﻠﻘﻴﻢ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻹﻧﺤ‬ ‫أﺣﺴﺐ‬ ‫ﺛﻢ‬ 9 ، 7 ، 5 ،3 ، 1 ‫اﻷﻋﺪاد‬ ‫ﻣﻦ‬ 1 ‫إﻃﺮح‬
‫؟‬ ‫ﺗﻼﺣﻆ‬ ‫ﻣﺎذا‬ (14) ‫ﻣﺜﺎل‬ ‫ﻣﻊ‬
9 ، 7 ، 5 ،3 ، 1 ‫اﻷﻋﺪاد‬
8 ، 6 ، 4 ،2 ، 0 : 1 ‫أﻃﺮح‬
X = ‫ــــــــــــــــــــــــــــ‬ = ‫ـــــــــ‬ = 4
S = - ( x )2
S = ‫ــــــــ‬ - 16 = 24 - 16
‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻹﻧﺤ‬ .... 8 = 2 2
(14) ‫ﻣﺜﺎل‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ (1) ‫ﻃﺮح‬ ‫ﻗﺒﻞ‬ ‫ﻟﻸﻋﺪاد‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻹﻧﺤ‬ ‫ﻧﻔﺲ‬ ‫ﻧﻼﺣﻆ‬
‫اﳌﺠﻤﻮع‬‫اﳌﺠﻤﻮع‬
x x2
1 1
3 9
5 25
7 49
9 81
25 165
x x2
0 0
2 4
4 16
6 36
8 64
20 120

152
: Standard Degree ‫المعياريه‬ ‫الدرجه‬
(7 - 4) ‫ﺗﻌﺮﻳﻒ‬
‫ﻋﻦ‬ ‫اﻟﻤﺘﻐﻴﺮ‬ ‫ذﻟﻚ‬ ‫ﻗﻴﻤﺔ‬ ‫اف‬‫ﺮ‬‫إﻧﺤ‬ ‫ﻗﺴﻤﺔ‬ ‫ﺧﺎرج‬ ‫ﺑﺄﻧﻬﺎ‬ ‫اﻟﻤﻌﻴﺎرﻳﺔ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﺗﻌﺮف‬ : ‫اﻟﻤﻌﻴﺎرﻳﺔ‬ ‫اﻟﺪرﺟﺔ‬
.‫ﻟﻬﺎ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻟﺘﻠﻚ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬
:‫اﻟﻤﻌﻴﺎرﻳﺔ‬ ‫اﻟﺪرﺟﻪ‬ ‫أﻧﻪ‬ ‫أي‬
Correlation � ‫اﻻرﺗﺒﺎط‬ [ 7 � 5 � 3 �
(7 - 5) ‫ﺗﻌﺮﻳﻒ‬
‫إﻟﻰ‬ ‫اﻻﺧﺮ‬ ‫ﻳﻤﻴﻞ‬ ‫ﻣﻌﻴﻦ‬ ‫ﺑﺎﺗﺠﺎه‬ ‫اﺣﺪﻫﻤﺎ‬ ‫ﺗﻐﻴﺮ‬ ‫إذا‬ ‫ﺑﺤﻴﺚ‬ ،‫ﻣﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻟﺮﻳﺎﺿﻴﺔ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﻫﻮ‬ : ‫اﻻرﺗﺒﺎط‬
‫اﻟﺘﻐﻴﺮ‬ ‫ﻛﺎن‬ ‫إذا‬ ‫أﻣﺎ‬ ،ً‫ﺎ‬‫ﻃﺮدﻳ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺳﻤﻲ‬ ‫واﺣﺪ‬ ‫ﺑﺎﺗﺠﺎه‬ ‫اﻟﺘﻐﻴﺮ‬ ‫ﻛﺎن‬ ‫ﻓﺎذا‬ ،ً‫ﺎ‬‫أﻳﻀ‬ ‫ﻣﻌﻴﻦ‬ ‫إﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫اﻟﺘﻐﻴﻴﺮ‬
. ً‫ﺎ‬‫ﻋﻜﺴﻴ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺳﻤﻲ‬ ‫ﻣﺘﻌﺎﻛﺴﻴﻦ‬ ‫ﺑﺎﺗﺠﺎﻫﻴﻦ‬
: x ، y ‫المتﻐيرين‬ ‫بين‬ ( ) Correlation Cofficient ‫االرتباط‬ ‫معامل‬
‫ﺣﻴﺚ‬ r :‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﻧﺤﺴﺐ‬
x ‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ = x ‫ﺣﻴﺚ‬
y ‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ = y
x ‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ =S x
y ‫ﻟﻠﻤﺘﻐﻴﺮ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ =S y
: (r) ‫ﺧصائﺺ‬ ‫بعﺾ‬
. (‫)اﻟﻤﻮﺟﺐ‬ ‫اﻟﻄﺮدي‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻣﻮﺟﺒﺔ‬ r (1)
.‫اﻟﺘﺎم‬ ‫اﻟﻄﺮدي‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ r=1 (2)
. (‫)اﻟﺴﺎﻟﺐ‬ ‫اﻟﻌﻜﺴﻲ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺳﺎﻟﺒﺔ‬ r (3)
. ‫اﻟﺘﺎم‬ ‫اﻟﻌﻜﺴﻲ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ r= -1 (4)
.‫اﻻرﺗﺒﺎط‬ ‫إﻧﻌﺪام‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ r= 0 (5)
‫ﻣﻦ‬ r ‫ﻗﻴﻤﺔ‬ ‫إﻗﺘﺮﺑﺖ‬ ‫وﻛﻠﻤـﺎ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫اﻻرﺗﺒــﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﻗﻴﻤﺔ‬ ‫أن‬ ‫ﺳﺒﻖ‬ ‫ﻣﻤﺎ‬ ‫ﻳﻼﺣﻆ‬
‫ﻫﺬا‬ ‫ﻛﺎن‬ ‫اﻟﺼﻔﺮ‬ ‫ﻣﻦ‬ ‫ﻗﻴﻤﺘﺔ‬ ‫إﻗﺘﺮﺑﺖ‬ ‫وﻛﻠﻤﺎ‬ ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻗﻮة‬ ‫ﻋﻠﻰ‬ ً‫ﻼ‬‫دﻟﻴ‬ ‫ﻫﺬا‬ ‫ﻛﺎن‬ 1 ‫أو‬ 1
. ‫اﻻرﺗﺒﺎط‬ ‫إﻧﻌﺪام‬ ‫ﻋﻠﻰ‬ ً‫ﻼ‬‫دﻟﻴ‬
+-
[ -1 , 1 ]
r= ‫ـــــــــــــــــــــــــــ‬
∑ x y
n
- x y
Sx
Sy
X - X
S
SD= ‫ـــــــــــــــ‬
r

153
� � 16 ‫ﻣﺜﺎل‬
: ‫ﻛﺎن‬ ‫إذا‬ x=5 ‫ﻟﻠﻌﺪد‬ ‫اﻟﻤﻌﻴﺎرﻳﺔ‬ ‫اﻟﺪرﺟﺔ‬ ‫ﺟﺪ‬ ‫ﺛﻢ‬ x، y ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﺟﺪ‬
‫ﻧﻮﻋﻪ؟‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬
S x
=
S y
=
‫ﺗﺎم‬ ‫ﻃﺮدي‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻧﻮع‬ ∴
x yy2
x2
yx
24121
816442
1836963
32641684
5010025105
110220553015
‫ﻧﻮﻋﻪ؟‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬
54321x
108642y
15
5
X =
Y = 30
5
55
5 -9 = 2
220
5
-36 = 8 = 2 2
=3
=6
‫اﳌﺠﻤﻮع‬
r = ‫ـــــــــــــــــــــــــــ‬ = ‫ـــــــــــــــــــــــــــــ‬
∑ x y
n - x y
Sx
Sy
= 1= 4
4
- (3)(6)
( 2) (2 2)
22 - 18
4
r= ‫ــــــــــــــــــــــ‬
110
5
SD =
x − x
Sx
∴SD =
5 − 3
2
=
2
2
= 2

154
� � 17 ‫ﻣﺜﺎل‬
: ‫ﻛﺎن‬ ‫إذا‬ ‫ﻧﻮﻋﻪ‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬ x ، y ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﺟﺪ‬
‫ــــــــــ‬ = 1
‫ــــــــــ‬ = 3
S x
=
S y
=
r = ‫ـــــــــــــــــــــــــــ‬ = ‫ــــــــــــــــــــــــــــ‬
‫ﺗﺎم‬ ‫ﻋﻜﺴﻲ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻧﻮع‬ r=
741-2-5x
-30369y
5
5
X =
Y = 15
5
95
5 - (1)2
= 19 -1 = 18 = 3 2
135
5 - (1)2
= 27 -9 = 18 = 3 2
∑ x y
n - x y
Sx
Sy
-75
15 - (1) (3)
3 2 × 3 2
= -1=
-15 -3 -18
(9) (2) (18)
xyy2
x2
yx
-4581259- 5
-123646-2
39131
001604
-21949-37
-7513595155

155
( 7 - 2 ) ‫تمرينات‬
/ 1‫س‬
3 ،0 ،8 ،7 ،9 ،12 :‫اﻟﺘﺎﻟﻴﺔ‬ ‫ﻟﻠﻘﻴﻢ‬ ‫اﻟﻤﺪى‬ ‫أوﺟﺪ‬ ( ‫أ‬
: ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻦ‬ ‫اﻟﻤﺪى‬ ‫أوﺟﺪ‬ (‫ب‬
/ 2‫س‬
10 ، 8 ،6 ،4 ،2 : ‫اﻟﺘﺎﻟﻴﺔ‬ ‫ﻟﻠﻘﻴﻢ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫إﺣﺴﺐ‬ ‫ﺛﻢ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫ﻋﺮف‬
/ 3‫س‬
‫ﻫﺬه‬ ‫أن‬ ‫وأﺛﺒﺖ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻋﺪد‬ ‫ﻛﻞ‬ ‫اﻟﻰ‬ 5 ‫أﺿﻒ‬ ‫ﺛﻢ‬ 5 ،7 ،1 ،2 ،6 ،3 : ‫ﻟﻸﻋﺪاد‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻻﻧﺤ‬ ‫أوﺟﺪ‬
. ‫اﻟﺤﺴﺎﺑﻲ‬ ‫اﻟﻮﺳﻂ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺆﺛﺮ‬ ‫وﻟﻜﻨﻬﺎ‬ ‫اﻟﻤﻌﻴﺎري‬ ‫اف‬‫ﺮ‬‫اﻹﻧﺤ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺆﺛﺮ‬ ‫ﻻ‬ ‫اﻹﺿﺎﻓﺔ‬
/ 4‫س‬
‫؟‬ ‫ﻧﻮﻋﻪ‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬ x، y ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﺟﺪ‬
/ 5‫س‬
‫ﻟﻠﻘﻴﻢ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫،ﺟﺪ‬ ‫آﺧﺮ‬ ‫ﺟﺪول‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺼﻞ‬ 4 ‫ﻓﻲ‬ x ‫ﻗﻴﻢ‬ ‫ﺿﺮﺑﺖ‬ ‫ﻟﻮ‬ ‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺴﺆال‬ ‫ﻓﻲ‬
. ‫اﻟﺴﺎﺑﻖ‬ ‫ﺑﺎﻟﺴﺆال‬ ‫اﻟﻨﺘﻴﺠﺔ‬ ‫وﻗﺎرن‬ ‫اﻟﺠﺪﻳﺪة‬
/ 6‫س‬
‫ﻧﻮﻋﻪ؟‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬ x ، y ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﺟﺪ‬
30 - 322826242220‫اﻟﻌﻤﺮ‬ ‫ﻓﺌﺎت‬
251020105‫ار‬‫ﺮ‬‫اﻟﺘﻜ‬
‫ﻟﻠﻘﻴﻢ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫،ﺟﺪ‬ ‫آﺧﺮ‬ ‫ﺟﺪول‬ ‫ﻋﻠﻰ‬ ‫ﺗﺤﺼﻞ‬
321x
642y
‫ﻧﻮﻋﻪ؟‬ ‫ﺑﻴﻦ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺘﻐﻴﺮﻳﻦ‬ ‫ﺑﻴﻦ‬ ‫اﻻرﺗﺒﺎط‬ ‫ﻣﻌﺎﻣﻞ‬ ‫ﺟﺪ‬
1284x
642y
31-5-9-13-x
-5-3-1+1+3y
_ _ _ _ _

156
‫اﻟﻤﻘﺪﻣﺔ‬
‫اﻻول‬ ‫اﻟﺠﺰء‬
‫اﻟﺮﻳﺎﺿﻲ‬ ‫اﻟﻤﻨﻄﻖ‬ : ‫اﻷول‬ ‫اﻟﻔﺼﻞ‬
‫واﻟﻤﺘﺒﺎﻳﻨﺎت‬ ‫اﻟﻤﻌﺎدﻻت‬ : ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻔﺼﻞ‬
‫واﻟﺠﺬور‬ ‫اﻻﺳﺲ‬ :‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﻔﺼﻞ‬
‫اﻟﻤﺜﻠﺜﺎت‬ ‫ﺣﺴﺎب‬ :‫اﺑﻊ‬‫ﺮ‬‫اﻟ‬ ‫اﻟﻔﺼﻞ‬
‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﺠﺰء‬
‫اﻟﻤﺘﺠﻬﺎت‬ :‫اﻟﺨﺎﻣﺲ‬ ‫اﻟﻔﺼﻞ‬
‫اﻻﺣﺪاﺛﻴﺔ‬ ‫اﻟﻬﻨﺪﺳﺔ‬ :‫اﻟﺴﺎدس‬ ‫اﻟﻔﺼﻞ‬
‫اﻻﺣﺼﺎء‬ :‫اﻟﺴﺎﺑﻊ‬ ‫اﻟﻔﺼﻞ‬
‫اﻟﻤﺤﺘﻮﻳﺎت‬
‫اﻟﻤﺤﺘﻮﻳﺎت‬
3
4
21
40
58
89
109
135
156

الرياضيــــــــــــات للصف الرابع العلمي

More Related Content

What's hot

Viewers also liked

More from Ayad Haris Beden

الرياضيــــــــــــات للصف الرابع العلمي