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19 tháng 7 2019

A= a2(1-b2) - b(1-b) - a(1-b)

=a2(1-b)(1+b) - b(1-b) - a(1-b)

=(1-b)(a2+a2b-b-a)

=(1-b)[a(a-1) + b(a2-1)]= (1-b)[a(a-1) + b(a-1)(a+1) =(1-b)(a-1)(a+ab+b)

Chỗ nào ko hiểu thì hỏi nhé ^^

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

\(=2\left(4-x\right)\left(2x+1\right)\)

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)

\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)

\(=\left(3x-2\right)\left(3x-6\right)\)

\(=3\left(3x-2\right)\left(x-2\right)\)

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-3y+4\right)\)

23 tháng 10 2021

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

15 tháng 7 2021

`a)x^4+2x^2y+y^2`

`=(x^2+y)^2`

`b)(2a+b)^2-(2b+a)^2`

`=(2a+b-2b-a)(2a+b+2b+a)`

`=(a-b)(3a+3b)`

`=3(a-b)(a+b)`

`c)8a^3-27b^3-2a(4a^2-9b^2)`

`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`

`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`

`=9b^2(2a-3b)`

a) Ta có: \(x^4+2x^2y+y^2\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)

\(=\left(x^2+y\right)^2\)

b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)

\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)

\(=\left(a-b\right)\left(3a+3b\right)\)

\(=3\left(a+b\right)\left(a-b\right)\)

28 tháng 12 2021

\(a,15a^2b^3+5a^3b^2=5a^2b^2\left(3b+a\right)\\ b,x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

28 tháng 12 2021

a) 15a2b3+5a3b2=5a2b2(3b+a)
b) x2-2x+x-y2=( x2-y2)-(2x+x)
                     =(x-y)(x+y)-x(2-1)
                      =(x-y)(x+y)-x3

a: \(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)

\(=\left(2x-1\right)\left(-3x+2\right)\)

b: \(2x^2-5x+2\)

\(=2x^2-4x-x+2\)

\(=2x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(2x-1\right)\)

a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)