jawabanmanipulasi perpangkatan bentuk aljabar
TRANSCRIPT
MATEMATIKARIAMATEMATIKARIA
SOLUTIONS
RememberX2 + Y2 = (X+Y)2 – 2XY
X3 + Y3 = (X+Y)3 – 3XY(X+Y)
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
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
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
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