=(x²+x)-(xy+y)

=x(x+1)-y(x+1)

=(x-y)(x+1)

Phân tích thành nhân tử

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

\(=x.\left(x-y\right)+\left(x-y\right)\)

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

\(3x^2-3y^2-x+y\\ =3\left(x^2-y^2\right)-\left(x-y\right)\\ =3\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left[3\left(x+y\right)-1\right]\\ =\left(x-y\right)\left(3x+3y-1\right)\)

\(3x^2-3y^2-x+y=3\left(x^2-y^2\right)-\left(x-y\right)=3\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left[3\left(x+y\right)-1\right]=\left(x-y\right)\left(3x+3y-1\right)\)

\(x^3+y^3+x+y\)

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

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

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

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

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

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

help

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

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

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

1, (x-y)(x-3)

2,(x-y)(x+2)

3,(x+1)(x+5y)

`a^3-a^2 x -ay+xy`

`=a^2(a-x)-y(a-x)`

`=(a-x)(x^2-y)`

`x^2-2xy+x-2y`

`= (x^2+x)-(2xy+2y)`

`=x(x+1)-2y(x+1)`

`=(x+1)(x-2y)`

`x^2-2x+2y-xy`

`=x(x-2) + y(2-x)`

`=x(x-2)-y(x-2)`

`=(x-2)(x-y)`

1,(a^2-y)(a-x)

2,(x+1)(x-2y)