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

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

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

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

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

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

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

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

b) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

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

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

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

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

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

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

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

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

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

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

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

Ta có: x² + y² - x²y² + xy - x - y

       = (x² - x²y²) + (y² - y) + (xy - x)

       = - x²(y² - 1) + y(y - 1) + x(y - 1)

       = - x²(y + 1)(y - 1) + (y - 1)(x + y)

       = (y - 1)(x + y - x²y - x²)

       = (y - 1)[- (x² - x) - (x²y - y)]

       = - (y - 1)[x(x - 1) + y(x² - 1)]

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

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

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

Ta có: x² + y² - x²y² + xy - x - y

       = (x² - x²y²) + (y² - y) + (xy - x)

       = - x²(y² - 1) + y(y - 1) + x(y - 1)

       = - x²(y + 1)(y - 1) + (y - 1)(x + y)

       = (y - 1)(x + y - x²y - x²)

       = (y - 1)[- (x² - x) - (x²y - y)]

       = - (y - 1)[x(x - 1) + y(x² - 1)]

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

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

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

Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 59 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng

a.

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

b.

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

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

c.

\(x^2-12x+36-64y^2=\left(x-6\right)^2-\left(8y\right)^2\)

\(=\left(x-6+8y\right)\left(x-6-8y\right)\)

d.

\(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)\)

\(=\left(x+2\right)\left(x+5\right)\)

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