a: \(x^4-2x^3+x^2-2x\)

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

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

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

b: \(x^4+x^3-8x-8\)

\(=\left(x^4+x^3\right)-\left(8x+8\right)\)

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3-8\right)\)

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

x^3+x^2-2x-8

= (x-2)(x^2+3x+4)

nah bạn chúc bạn học tốt nha 

     x³ + x² - 2x - 8

=  ( x³ - 8 ) + ( x² - 2x )

=  ( x - 2 ) . ( x² + 2x + 4 ) + x ( x - 2 )

= ( x - 2 ) .( x² + 2x + 4 + x )

= ( x-2 ) . ( x² + 3x + 4 ) 

x^3+x^2-2x-8

= (  x^3 - 8) + ( x^2 - 2x )

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

=(  x - 2)( x^2 + 2x + 4 + x)  

=(  x - 2)( x^2 + 3x + 4 )  

\(x^3-8+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4+2x\right)=\left(x-2\right)\left(x^2+4x+4\right)\\ =\left(x-2\right)\left(x+2\right)^2\)

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)

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

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

=2(2x+1)(x-3)

3: \(=2\left(x+2\right)\left(25x-15-x\right)\)

\(=2\left(x+2\right)\left(24x-15\right)\)

=6(x+2)(8x-5)

\(x^2+2x-8\)

\(=x^2+4x-2x-8\)

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

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

\(x^2+5x+6\)

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

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

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

\(4x^2-12x+8\)

\(=4x^2-4x-8x+8\)

\(=4x\left(x-1\right)-8\left(x-1\right)\)

\(=\left(4x-8\right)\left(x-1\right)\)

\(x^2-xy-\dfrac{3}{4}y^2\)

\(=x^2-\dfrac{3}{2}xy+\dfrac{1}{2}xy-\dfrac{3}{4}y^2\)

\(=x\left(x-\dfrac{3}{2}y\right)+\dfrac{1}{2}y\left(x-\dfrac{3}{2}y\right)\)

\(=\left(x+\dfrac{1}{2}y\right)\left(x-\dfrac{3}{2}y\right)\)

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

=(x^4+1)^2-x^4

=(x^4-x^2+1)(x^4+x^2+1)

=(x^4-x^2+1)(x^4+2x^2+1-x^2)

=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)

c: =(x+1)(x^2-x+1)+2x(x+1)

=(x+1)(x^2-x+1+2x)

=(x+1)(x^2+x+1)

d: =(x^2-1)(x^2+1)-2x(x^2-1)

=(x^2-1)(x^2-2x+1)

=(x-1)^2*(x-1)(x+1)

=(x+1)(x-1)^3

Đặt \(A=\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8\). Rút gọn A,ta được:

\(A=x^4-4x^3+12x^2-16x+7\)

\(=x^4-2x^3+x^2-2x^3+4x^2-2x+7x^2-14x+7\)

\(=x^2\left(x^2-2x+1\right)-2x\left(x^2-2x+1\right)+7\left(x^2-2x+1\right)\)

\(=\left(x^2-2x+1\right)\left(x^2-2x+7\right)\)

\(=\left(x-1\right)^2\left(x^2-2x+7\right)\)

Ok chứ?