Jawabanmanipulasi Perpangkatan Bentuk Aljabar

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Jawabanmanipulasi Perpangkatan Bentuk Aljabar

  1. 1. MATEMATIKARIA
  2. 2. SOLUTIONS Remember X 2 + Y 2 = (X+Y) 2 – 2XY X 3 + Y 3 = (X+Y) 3 – 3XY(X+Y)
  3. 3. Jawaban Soal 1 a b c L = ½ ab = 270 ab = 540 a + b + c = 90 a + b = 90 - c a 2 + b 2 = c 2  (a + b) 2 - 2 ab = c 2 (90-c) 2 – 2.540 = c 2 8100-180c+c 2 – 1080 = c 2 180c =8100-1080 = 7020 c = 39
  4. 4. Soal 2 <ul><li>a + b + c = 0  a+b = -c </li></ul><ul><li>a 3 + b 3 + c 3 = 333 </li></ul><ul><li>(a+b) 3 – 3ab(a+b) + c 3 = 333 </li></ul><ul><li>(-c) 3 – 3ab(-c) + c 3 = 333 </li></ul><ul><li>-c 3 + 3abc + c 3 = 333 </li></ul><ul><li>3abc = 333 </li></ul><ul><li>abc = 111 </li></ul>
  5. 5. Soal 4 <ul><li>ab = a b </li></ul><ul><li>a/b = a 3b x </li></ul><ul><li>a 2 = a 4b </li></ul><ul><li>4b = 2  b = ½ </li></ul><ul><li>2a = a 3/2 </li></ul><ul><li>2a – a 3/2 = 0  a(2- a 1/2 )= 0 </li></ul><ul><li> a=0 or 2 – a 1/2 = 0 </li></ul><ul><li>a 1/2 = 2 </li></ul><ul><li>a = 4 </li></ul>
  6. 6. Soal 5 A – B = 3 (A – B) 3 = 3 3 A 3 – 3AB(A-B) – B 3 = 27 A 3 – 3AB.3 – B 3 = 27 A 3 – 9AB – B 3 = 27

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