Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(43x^3y^3-32x^2y^2\)
\(=x^2y^2\left(43xy-32\right)\)
b) \(ax-bx+ab-x^2\)
\(=\left(ax+ab\right)-\left(bx+x^2\right)\)
\(=a\left(b+x\right)-x\left(b+x\right)\)
\(=\left(a-x\right)\left(b+x\right)\)
c) \(12a^2b-18ab^2-30b^2\)
\(=6b\left(2a^2-3ab-5b\right)\)
d) \(27a^2\left(b-1\right)-9a^3\left(1-b\right)\)
\(=27a^2\left(b-1\right)+9a^3\left(b-1\right)\)
\(=\left(27a^2+9a^3\right)\left(b-1\right)\)
\(=9a^2\left(b-1\right)\left(a+3\right)\)
1. \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
2.\(8x^6-27y^3=\left(2x\right)^3-\left(3y\right)^3=\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(8x^6+6xy+27y^3\right)\)
3.\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)4.câu cuối là \(8b^3\)bạn nhé !!
tik mik nhé
k ) \(125x^3-1\)
\(=\left(5x\right)^3-1\)
\(=\left(5x-1\right)\left[\left(5x\right)^2+5x.1+1^2\right]\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m ) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right).\left[\left(x^2\right)^2+x^2.y+y^2\right]=\left(x^2-y\right).\left(x^4+x^2y+y^2\right)\)
n ) \(a^4-2a^2+1\)
\(=\left(a^2\right)^2-2.a^2.1+1^2=\left(a^2-1\right)^2\)
i ) \(a^3+6a^2+12a+8\)
\(=\left(a+2\right)^3\)
k) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
n) \(a^4-2a^2+1=\left(a^2-1\right)^2=\left(a^2-1\right)\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
i) \(a^3+6a^2+12a+8=\left(a+2\right)^2\)
đây là hằng đẳng thức
\(a^3+6a^2+12a+8=a^3+3.2.a^2+3.2^2.a+2^3=\left(a+2\right)^3\)
Đề là phân tích thành nhân tử ah ? mik làm vài câu thôi , mấy cau sau tương tự ah
=====================
a , x2 - 2ax -b2 +a2 = (x-a)2 - b2 = (x-a-b)(x-a+b)
b , x3(x2-y)2 - 36x = x{[x(x2-y)]2-62} = x[x(x2-y)+6][x(x2-y)-6]
1) 9(a+b)2-4(a-2b)2
=[3(a+b)]2-[2(a-2b)]2=....
2) 8x3+27y3=(2x)3+(3y)3=...
3) (a+b)3-c3=(a+b-c).[(a+b)2+(a+b)c+c2]
4) x3+3x2+3x+1=(x+1)3
a) \(6a^2b^2c-4ab^2c^2+12a^2bc^2\)
\(=2abc\left(3ab-2bc+6ac\right)\)
b)\(x^2\left(x-y\right)-y\left(y-x\right)\)
\(=x^2\left(x-y\right)+y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+y\right)\)
x9 + 1
= (x3)3 + 13
= (x3 + 1)(x6 - x3 + 1)
= (x + 1)(x2 - x + 1)(x6 - x3 +1)
8a3 - 12a2 + 6a - 1
= (2a)3 - 3(2a)21 + 3 . 2a . 12 - 1
= (2a - 1)3
27a3 - 54a2b + 36ab2 - 8b3
= (3a)3 - 3(3a)22b + 3 . 3a . (2b)2 - (2b)3
= (3a - 2b)3