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

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

`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)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

Đề sai rồi bạn

^2 + 4xy - 16 + 4y^2 
= x^2 + 4xy + 4y^2 - 4^2 
= (x + 2y)^2 - 4^2 
= (x + 2y - 4)(x + 2y + 4) 
2x^2-5xy-3y^2 
= 2^x + xy - 6xy - 3y^2 
= x(2x + y) - 3y(2x + y) 
= (2x + y)(x - 3y)

mình chưa hiểu

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