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: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

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

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

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

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

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

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

Gọi a+b =x có:  

a(x+b)³−b(x+a)³

 =a(x³+3x²b+3xb²+b³)−b(x³+3x²a+3xa²+a³)  

=ax³+3ax²b+3axb²+ab³−bx³−3bx²a−3bxa²−ba³  

=(a−b)x³+(3ax²b−3bx²a)+(3axb²−3bxa²)+ab³−ba³  

=(a−b)x³+3axb(b−a)+ab(b²−a²)  

=−x³(b−a)+3axb(b−a)+ab(b+a)(b−a)

 =−x³(b−a)+3axb(b−a)+(a²b+ab²)(b−a)

 =(b−a)(−x³+3axb+a²b+ab²)

nho lik e

\(a,x\left(a-b\right)+2a-2b=x\left(a-b\right)+2\left(a-b\right)=\left(a-b\right)\left(x+2\right)\\ b,Sửa:ax+ay+5x+5y=a\left(x+y\right)+5\left(x+y\right)=\left(a+5\right)\left(x+y\right)\)

\(a,=\left(x+2\right)\left(a-b\right)\\ b,Sửa:ax+ay+5x+5y\\ =a\left(x+y\right)+5\left(x+y\right)\\ =\left(a+5\right)\left(x+y\right)\)

a) \(x\left(a-b\right)+2a-2b\)

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

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

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

Bài 4: 

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

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

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

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

a(a+2b)3 -b(2a+b)3

\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left[\left(a^2\right)^2+ \left(b^2\right)^2\right]-2ab\left(a^2-b^2\right)\)

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

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

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

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