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

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)

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

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

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

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

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

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

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

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

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!