Your SlideShare is downloading. ×
Korenovanje
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
4,708
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
73
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. KORENOVANJENeka je a realan i n prirodan broj.Svako rešenje jednačine xn = a‘’po x’’ (ako postoji) naziva se n -ti koren broja a u oznaci x = n a .Dakle: simbol n a označava: 1) n − ti koren realnog broja a u svim slučajevima kada je on jedinstven (n ∈ N , n = 2k − 1, k ∈ N , a ∈ R) 2) Pozitivan n -ti koren broja a u slučaju n = 2k , k ∈ N , a > 0 Ova definicija sigurno nije baš mnogo jasna!!! Ajde da vidimo par primera: 3 1 3 ⎛1⎞ 1 3 27 = 3 = 3;3 3 3 = ⎜ ⎟ = 8 ⎝2⎠ 2 − 32 = 5 (−2)5 = −2; 5 7 0 =0 _______________________________________ 4 1 4 ⎛1⎞ 1 2 4 = 2 = 2; 2 2 4 = ⎜ ⎟ = 16 ⎝2⎠ 2 Pazi: 4 16 = 4 2 4 = 2; − 4 16 = − 4 2 4 = −2 Pogrešno je pisati: 4 16 = ±2 ZAPAMTI!!!Važi: ⎧a , n − neparann an = ⎨ ⎩ a, n − paranPrimeri: (pazi, dogovor je da je A = 2 A , to jest, jedino se ovde ne piše broj 2) 9 = 32 ; 3 23 = 2 (−3) 2 = − 3 = 3 ; 3 (−2) 3 = −2 www.matematiranje.com 1
  • 2. ZAPAMTI: Kad vidiš 2, 4, 6,… (parni koren) iz nekog konkretnog broja, rešenje je uvekpozitivan broj. Kad vadiš 3, 5, 7… (neparan koren) iz nekog broja, rešenje može biti inegativan broj, u zavisnosti kakva je potkorena veličina. (−5) 2 = − 5 = 5 3 53 = 5 4 ( −7 ) 4 = − 7 = 7 3 (−5) 3 = −5 5 ⎛1⎞ 1 6 (−12) = − 12 = 12 6 5 ⎜ ⎟ = ⎝ 10 ⎠ 10 8 7 ⎛ 1⎞ 1 1 ⎛ 3⎞ 3 8 ⎜− ⎟ = − = 7 ⎜− ⎟ = − ⎝ 3⎠ 3 3 ⎝ 5⎠ 5Primer: Za koje realne brojeve x je tačna vrednost: a) x2 = x b) 3 x3 = − x v) x2 = −x g) x 4 = x ( ) 4____________________________________Rešenje: a) x 2 = x je tačna samo za vrednosti x koje su veće ili jednake nuli, jer ⎧ x, x>0 ⎪ Dakle x ≥ 0 x = ⎨− x , 2 x<0 ⎪0, za ,x = 0 ⎩b) 3 x 3 = − x je tačka samo za x = 0 !!! Zašto? Ako uzmemo da je x negativan broj, naprimer x = −5 ⇒ 3 (−5) 3 = −5 ≠ − (−5) = +5 , a ako uzmemo x > 0 , recimo x = 103 103 = 10 ≠ −10v) x 2 = − x , x mora biti manje od nule, ili nula jer kao malopre važi: ⎧ x, x > 0 ⎪ x = ⎨− x, x < 0 , Dakle x ≤ 0 2 ⎪0, x = 0 ⎩ www.matematiranje.com 2
  • 3. ( ) 4g) x 4 = x , Ovde mora biti x ≥ 0 . Zašto? Zbog ( x)4 koji ne može biti negativanodnosno mora biti x > 0Pravila: m 1) n am = a n 2) n a ⋅b = n a ⋅ n b 3) n a : n b = n a :b 4) ( a) n m = n am 5) n m a = n⋅m a a mp = n a m ( p se skrati) np 6)Moramo naglasiti da pravila važe pod uslovima da je: a, b → pozitivni realni brojevim, n, p → prirodni brojevi. Zadaci:1) a) Izračunaj 36 − 2 25 + 4 16 − 5 32 36 − 2 25 + 4 16 − 5 32 = = 6 − 2 ⋅ 5 + 4 2 4 − 5 25 = = 6 − 10 + 2 − 2 = −4 9 3 1 4 b) Izračunaj + + 16 4 8 2 3 ⎛3⎞ ⎛1⎞ = ⎜ ⎟ + 3 ⎜ ⎟ + 4 24 = ⎝2⎠ ⎝2⎠ 3 1 + +2=4 2 2 2 ⎛4⎞ 3 c) Izračunaj ⎜ ⎟ + − 27 − 4 ⎝9⎠ 2 ⎛4⎞ 3 ⎜ ⎟ + − 27 − 4 = ⎝9⎠ 2 ⎛2⎞ 2 2 1 = ⎜ ⎟ + 3 (−3) 3 − 2 = − 3 − 2 = − 5 = −4 ⎝3⎠ 3 3 3 www.matematiranje.com 3
  • 4. d) Izračunaj 9 ⋅ 3 (−8) ⋅ 5 − 32 9 ⋅ 3 (−8) ⋅ 5 − 32 = = 32 ⋅ 3 (−2) 3 ⋅ 5 (−2) 5 = = 3 ⋅ (−2) ⋅ (−2) = 122) Izračunaj ( x − 5) 2 + ( x + 5) 2 ( x − 5) 2 + ( x + 5) 2 = x − 5 + x + 5Kako je: ⎧ x − 5, za x − 5 ≥ 0 ⎧ x − 5, za x ≥ 5 x−5 = ⎨ = ⎨ i ⎩− ( x − 5), za x − 5 < 0 ⎩− ( x − 5), za x < 5 ⎧ x + 5, za x + 5 ≥ 0 ⎧ x + 5, za x ≥ −5 x+5 = ⎨ = ⎨ ⎩− ( x + 5), za x + 5 < 0 ⎩− ( x + 5), za x < −5moramo najpre videti ‘’ gde ima’’ rešenja I za x ≥ 5 i x ≥ −5 x − 5 + x − 5 = x − 5 + x + 5 = 2x x ∈ [5, ∞ ) __________________________________________________________________ II za x ≥ 5 i x < −5 Nema rešenja __________________________________________________________________ www.matematiranje.com 4
  • 5. III za x < 5 i x ≥ −5 x − 5 + x − 5 = − x + 5 + x + 5 = +10 x ∈ [− 5,5) ________________________________________________________________ IV za x<5 i x < −5 x − 5 + x + 5 = − x + 5 − x − 5 = −2 x ⎧− 2 x, x < −5 ⎪ Konačno: x − 5 + x + 5 = ⎨10, − 5 ≤ x < 5 ⎪2 x, x ≥ 5 ⎩RacionalisanjeKoristeći se osobinama korena, možemo, u cilju uprošćavanja nekog izraza, odstranitikorene ili ih premestiti na željeno mesto. Primeri: 1 1 5 5 5 1) = ⋅ = = 5 5 5 5 2 5 9 9 12 9 12 3 12 3 4 ⋅ 3 3 ⋅ 2 3 3 3 2) = ⋅ = = = = = 12 12 12 12 4 4 4 2 15 15 3 15 3 5 3 3) = ⋅ = = 2 3 2 3 3 2⋅3 2Ovo je bio najprostiji tip zadataka. Gde racionalisanjem prebacujemo koren iz imenioca ubrojilac. nU sledecoj grupi zadataka ćemo koristiti da je n a =a, a >0 www.matematiranje.com 5
  • 6. 64) 3 = da bi ‘’uništili ’’ koren u imeniocu moramo napraviti 3 23 , a pošto imamo 2 6 3 treba racionalisati sa 2 2 . Dakle:3 1 2 6 6 3 22 6 ⋅ 3 22 6 ⋅ 3 4 =3 ⋅ = = = 33 43 2 2 2 3 2 3 2 3 2 10 10 4 33 10 4 27 10 4 275) = ⋅ = = = 4 3 4 3 4 33 4 4 3 3 3 ab ab a1b 2 ab 3 ab 2 ab 3 ab 2 3 26) = ⋅ = = = ab 3 a 2b 3 a 2b1 3 a1b 2 3 a 3b 3 abKad u imeniocu imamo zbir ili razliku dva kvadratna korena, upotrebljavamo razlikukvadrata: ( A − B) ⋅ ( A + B) = A2 − B 2 1 1 2− 3 2− 3 2− 37) = ⋅ = = = 2− 3 2 + 3 2 + 3 2 − 3 22 − 3 2 4−38) 11 = 11 ⋅ 6 + 2 11 6 + 2 11 6 + 2 11 6 + 2 = = = ( ) ( ) ( ) 6− 2 6− 2 6+ 2 2 6 − 2 2 6−2 49) 5 = 5 ⋅ 2 3 +3 2 = 5 2 3+3 2 ( ) = ( ) ( ) ( 5 2 3 +3 2 5 2 3 +3 2 5 2 3 +3 2 = = ) 22 3 −3 2 2 3 −3 2 2 3 +3 2 2 3 − 3 2 ( ) ( ) 2 4⋅3 − 9⋅ 2 12 − 18 −6U zadacima u kojima se u imeniocu javlja zbir ili razlika ‘’trećih’’ korena moramokoristiti: A3 − B 3 = ( A − B)( A2 + AB + B 2 ) → Razlika kubova A3 + B 3 = ( A + B)( A2 − AB + B 2 ) → Zbir kubova10) 1 1 3 − 3 3 3 2 + 3 22 3 9 − 3 6 + 3 4 3 9 − 3 6 + 3 4 3 9 − 3 6 + 3 4 3 2 =3 ⋅ = = =3 3+3 2 3 + 3 2 3 32 − 3 3 3 2 + 3 2 2 3 3 3 + 2 3 3 3+ 2 5________________________________________________________________________11) 5 =3 5 ⋅ 3 52 + 3 5 3 4 + 3 4 2 5 3 25 + 3 20 + 3 16 = ( = 5 3 25 + 3 20 + 3 16 ) ( ) 3 33 5− 4 3 5− 4 5 + 5 4+ 4 3 3 2 3 3 3 2 3 5 − 43________________________________________________________________________ www.matematiranje.com 6
  • 7. 312) = ovde ćemo uraditi dupli racionalizaciju da bi uništili četvrti koren. 4 5 −2 3 =4 3 4 ⋅4 5+2 3 5+2 = 4 = 3 4 5+2 ⋅ 5+4 3 = ( ) ( ) ( 4 5+2 )( 5+4 ) = 3( 4 5+2 )( 5+4 )4 5−2 5 − 2 5 + 2 4 52 − 2 2 5−4 5+4 52 − 4 2 −11Vratimo se na zadatke sa korenima:3) Izračunati: a) 3 x 2 x −1 ⋅ 3 x −1 x b) x3 x 2 ⋅ 3 x 2 : (x ) −1 3Rešenje: a) 2 1 2 1 1 1 − − −3 −1 −1 2 3 −1 −1 5 −1 −1 −1 x 2 x ⋅ x 5 x = x 3 x ⋅ x 5 x = x ⋅ x ⋅x ⋅ x 3 6 5 10 = x ⋅x ⋅x ⋅x = 3 6 5 10 2 1 1 1 20 −5− 6 +3 12 2 − − +=x 3 6 5 10 =x 30 =x 30 = x = 5 x25 (x )= 3 2 −1 x3 x 2 ⋅ 3 x 2 : x 3 x 2 ⋅ x 3 : x −3 = b) 1 2 2 3 1 2 2 ⎛ 3⎞ 3+ 2 + 4 + 9 18 − + + −⎜ − ⎟ x ⋅x ⋅x :x 2 6 3 2 =x 2 6 3 ⎝ 2⎠ =x 6 = x = x3 6 1ZAPAMTI: x = x24) Izračunaj: a) 5 2 + 3 8 − 50 − 98 b) 3 + 3 27 − 2 48Ovde je ideja da upotrebom pravila za korenovanje a ⋅ b = a ⋅ b , svaki sabiraksvedemo na čist koren. www.matematiranje.com 7
  • 8. a)5 2 + 3 8 − 50 − 98 =5 2 + 3 4 ⋅ 2 − 25 ⋅ 2 − 49 ⋅ 2 =5 2 + 3⋅ 2 2 − 5 2 − 7 2 =5 2 + 6 2 −5 2 − 7 2 = − 2b) 3 + 3 27 − 2 48 = 3 + 3 9 ⋅ 3 − 2 16 ⋅ 3 = 3 + 3⋅3 3 − 2 ⋅ 4 3 = 3 + 9 3 − 8 3 = 2 35) Izračunaj: 3 ⋅ 12 4 − 4 ⋅ 15 27 + 8 ⋅ 24 16 + 5 ⋅ 20 81Rešenje: 3 ⋅ 12 4 − 4 ⋅ 15 27 + 8 ⋅ 24 16 + 5 ⋅ 20 81 = 3 ⋅ 12 2 2 − 4 ⋅ 15 33 + 8 ⋅ 24 2 4 + 5 ⋅ 20 34 = 3 ⋅ 6 2 − 45 3 + 8 ⋅ 6 2 + 55 3 = −− −−− −− −−− = 11 2 + 3 6 5 LAGRANŽOV INDENTITET: a + a2 − b a − a2 − b a± b = ± 2 2Gde je a > 0, b > 0, b < a 2Primenimo ga na 2 primera: a) 2+ 3 b) 6−4 2 2 + 22 − 3 2 − 22 − 3a) 2+ 3 = + 2 2 2+ 1 2− 1 = + 2 2 3 1 3 +1 = + = 2 2 2 www.matematiranje.com 8
  • 9. b) 6 − 4 2 = pazi, prvo moramo 4 da ubacimo pod koren!!! 6 + 6 2 − 32 6 − 6 2 − 32 6 − 16 ⋅ 2 = 6 − 32 = + 2 2 6 + 36 − 32 6 − 36 − 32= + 2 2 6+2 6−2= + = 4 − 2 = 2− 2 2 2 2+ 3 2− 37) Dokazati da je vrednost izraza + iracionalan broj. 2 + 2+ 3 2 − 2− 3Najpre ćemo upotrebom Lagranžovog indetiteta ‘’ srediti’’ 2+ 3 i 2− 3 3 +1 3 −1 2 + 3 =(prethodni zadatak) = , slično je i 2− 3 = , Dakle: 2 2 2+ 3 2− 3 2+ 3 2− 3 + = + = 2 + 2+ 3 2 − 2− 3 3 +1 3 −1 2+ 2− 2 2 2+ 3 2− 3= + = Pazi na znak!!! 2 + 3 +1 2 − 3 +1 2 2= 2 2+ 3( + ) 2 2− 3 ( ) 3+ 3 3− 3= (2 )( ) ( 2 + 6 3− 3 + 2 2 − 6 3+ 3 )( ) ( 3+ 3 3− 3 )( ) 6 2 −2 6 +3 6 − 18 + 6 2 + 2 6 −3 6 − 18= 2 32 − 3 12 2 − 2 18 12 2 − 2 9 ⋅ 2 12 2 − 6 2 6 2= = = = = 2 9−3 6 6 6 www.matematiranje.com 9
  • 10. 4+2 3 8) Dokazati da je: = 3 +1. 10 + 6 3 3Poći ćemo od leve strane da dobijemo desnu.4 + 2 3 = 3 +1+ 2 3 = 3 + 2 3 +1 = ( 3) + 2 2 3 +1 = ( 3 +1 ) 210 + 6 3 = razmislimo da li ovo nije 10 + 6 3 = ( A + B ) ? 3( A + B) 3 = A3 + 3 A2 B + 3 AB 2 + B 3( ) 3 3 2 3 + 1 = 3 + 3 ⋅ 3 ⋅1 + 3 ⋅ 3 ⋅12 + 13 = 27 + 3 ⋅ 3 ⋅1 + 3 3 + 1 = 9 ⋅ 3 + 9 + 3 3 + 1 = 3 3 + 3 3 + 10 = 10 + 6 3Dakle: 4+2 3 = ( 3 + 1) = ( 2 3 +1 ) 2 = 3 +1 3 10 + 6 3 3 ( 3 + 1) 3 3 +1Ovim je dokaz završen!!! 6 9) Racionalisati: 21 + 7 + 2 3 + 2 6 6 6 = = 21 + 7 + 2 3 + 2 7 ⋅ 3 + 7 + 2 3 +1 7 3 +1 + 2 3 +1 ( ) ( ) ( ) 6 3 −1 7 − 2= ⋅ ⋅ = ( 3 +1 )( 7 +2 ) 3 −1 7 − 2= ( 6 3 −1 7 − 2 = )( 6 3 −1 7 − 2 = ) ( )( ) ( 3 −1 )( 7 −2 ) ⎛ 3 − 12 ⎞⎛ 7 − 2 2 ⎞ ⎜ 2 ⎟⎜ 2 ⎟ 2⋅3 ⎝ ⎠⎝ ⎠ www.matematiranje.com 10
  • 11. 110) Racionalisati: 3 9 + 6 +3 4 3 1 1 3 3−3 2 3 3−3 2 3 3 −32 = ⋅3 = = =3 9 + 3 6 + 3 4 3 32 + 3 3 ⋅ 3 2 + 3 22 3 − 3 2 3 33 − 3 23 3− 2 3 3 −32= =33 −32 1Ovde smo imali A2 + AB + B 2 , pa smo dodali A-B, da bi dobili A3 − B 3 . www.matematiranje.com 11

×