jawabanmanipulasi perpangkatan bentuk aljabar

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MATEMATIKARIA MATEMATIKARIA

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Page 1: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

MATEMATIKARIAMATEMATIKARIA

Page 2: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

SOLUTIONS

RememberX2 + Y2 = (X+Y)2 – 2XY

X3 + Y3 = (X+Y)3 – 3XY(X+Y)

Page 3: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

Jawaban Soal 1Jawaban Soal 1

ab

c

L = ½ ab = 270 ab = 540a + b + c = 90a + b = 90 - c

a2 + b2 = c2 (a + b)2 - 2 ab = c2

(90-c)2 – 2.540 = c2

8100-180c+c2 – 1080 = c2

180c =8100-1080 = 7020c = 39

Page 4: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

Soal 2Soal 2

a + b + c = 0 a + b + c = 0 a+b = -c a+b = -c aa33 + b + b33 + c + c33 == 333333(a+b)(a+b)33 – 3ab(a+b) + c – 3ab(a+b) + c33 = 333= 333(-c)(-c)33 – 3ab(-c) + c – 3ab(-c) + c33 = 333= 333-c-c33 + 3abc + c + 3abc + c33 = 333= 333

3abc3abc = 333= 333abcabc = 111= 111

Page 5: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

Soal 4Soal 4ab = aab = abb

a/b = aa/b = a 3b3b xx

aa22 = a= a4b4b

4b = 2 4b = 2 b = ½ b = ½2a = a2a = a3/2 3/2

2a – a2a – a3/23/2 = 0 = 0 a(2- a a(2- a1/21/2)= 0)= 0 a=0 a=0 or or 2 – a2 – a1/21/2= 0= 0

aa1/2 1/2 = 2= 2aa = 4 = 4

Page 6: Jawabanmanipulasi Perpangkatan Bentuk Aljabar

Soal 5Soal 5399 33 =−−+ xx 33 9,9: −=+= xBxAMisal

A – B = 3

(A – B)3 = 33

A3– 3AB(A-B) – B3 = 27

A3 – 3AB.3 – B3 = 27

A3 – 9AB – B3 = 27 80181

)1(81

181

3812

2781918

27)9(9999

2

32

3 2

3 2

3 2

33

=−=

−=−

−=−

=−−

=−−

=−−−+−+

x

x

x

x

x

xxxx