2. \(2xy-4y+2y^2\)

 \(=2y\left(x-2+y\right)\)

Lời giải:

$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$

$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$

a: =(x-3)(x-1)(x+1)

A

a/ \(x^3-3x^2-x+3\)

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

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

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

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

b/ \(x^2y-2xy^2+y^3\)

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

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

= y-2xy\(^2\)-x\(^2\)y-y\(^3\)= y(1-2xy)-y(x\(^2\)+y\(^2\))= y(1-2xy-x\(^2\)-y\(^2\))= y(-x\(^2\)-2xy-y\(^2\)+1)

=-y(x\(^2\)+2xy+y\(^2\)-1)= -y[(x+y)\(^2\)-1] = -y(x+y-1)(x+y+1)

a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)

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

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

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

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

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

\(2x^2y+4xy^2+2y^3-8\) 

\(=2y\left(x^2+2xy+y^2\right)-8\)

\(=2y\left(x+y\right)^2-8\) 

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

thank you huỳnh thanh phong