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

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

e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

e)x³-4x+14x(x-2)=0

⇔ x(x²-4)+14x(x-2)=0

⇔ x(x-2)(x+2)+14x(x-2)=0

⇔ (x-2)(x²+2x+14x)=0

⇔ x(x-2)(x+16)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)

g)x²(x+1)-x(x+1)+x(x-1)=0

⇔ (x+1)(x²-x)+x(x-1)=0

⇔ x(x+1)(x-1)+x(x-1)=0

⇔ x(x-1)(x+2)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)

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

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

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

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

x(x-y)² -y(x-y)²+xy²-x²y=x(x-y)² -y(x-y)²+(xy²-x²y)=x(x-y)² -y(x-y)²+xy(x-y)=\(\left(x-y\right)\left[x\left(x-y\right)-y\left(x-y\right)+xy\right]\)=\(\left(x-y\right)\left[\left(x-y\right)^2+xy\right]\)

 2(x-y)² -y(x-y)² +xy²-x²y= 2(x-y)²-y(x-y)²+(xy^2-x^2y)=2(x-y)²-y(x-y)²+xy(x-y)=(x-y)\(\left[2\left(x-y\right)-y\left(x-y\right)+xy\right]\)=(x-y)(2x-2y-xy+y²+xy)=(x-y)(2x-2y+y²)

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

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

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

\(=\left(x-y\right)\left[\left(x-y\right)\left(2-y\right)-xy\right]\)

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

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

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

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

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/ \(x^2y-2xy^2+y^3\)

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

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