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

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

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

(x-2)(x+y)

bạn rút x ra rồi tính ❕

= ( x³ + 2x²y + xy² ) - 16x

= x (x² + 2xy + y²) - 16x

= x( x + y)² - 16x

= x [ ( x + y)² - 16 ]

= x ( x + y +4) ( x + y - 4)

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

1.A

2.C

3.B

4.C

a

c

b

c

^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

Ý a có rì đó sai sai nha bn 

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

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

x²-xy+2x-2y
= (x²-xy)+(2x-2y)
= x(x-y)+2(x-y)
= (x-y)(x+2)

ax+ay-2x-2y
= (ax+ay)-(2x+2y)
= a(x+y)-2(x+y)
=(x+y)(a-2)

ax²-3axy+bx-3by
= (ax²+bx)-(3axy+3by)
= x(ax+b)-3y(ax+b)
= (ax+b)(x-3y)

2a²-5by-5a²y+2bx
= (2a²-5a²y)+(2bx-5by)
= a²(2-5y)+b(2x-5y)