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

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

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

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

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

x³ + 2x²y + xy² - 4x

= x( x² + 2xy + y² - 4 )

= x[ ( x + y )² - 2² ]

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

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

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

\(=x\left(x+y\right)^2-4x=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\)

a) x^ 3 + 2x^ 2y + xy ^2 − 4x

= x (x^ 2 + 2xy + y^ 2 − 4)

= x [(x + y)^ 2 − 2 ^2]

= x (x + y + 2)(x + y − 2) 

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

\(x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

\(=x\left[\left(x^2+2xy+y^2\right)-9\right]\)

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

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

? đây là phần hỏi bài mà....

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)

=x.(x²+2xy+y²-16)

=x.[(x+y)²-16]

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

sửa đề câu a đi 

\(x^2-6x-7=x^2+x-7\left(x+1\right)=\left(x-7\right)\left(x+1\right)\)

+)\(x^3+2x^2+xy^2-4x\)

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

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

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

+) \(x^2-6x-7\)

\(=x^2-6x+9-16\)

\(=\left(x-3\right)^2-16\)

\(=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)

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

\(x^2-6x-7=x\left(x+1\right)-7\left(x+1\right)=\left(x-7\right)\left(x+1\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)