xy(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

 xy(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

.

.

.

 xy(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

\(xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)\)

\(=xy\left(x-y\right)+yz\left[\left(y-x\right)-\left(z-x\right)\right]+xz\left(z-x\right)\)

\(=xy\left(x-y\right)-yz\left(x-y\right)-yz\left(z-x\right)+xz\left(z-x\right)\)

\(=\left(x-y\right)\left(xy-yz\right)-\left(z-x\right)\left(yz-xz\right)\)

\(=\left(x-y\right)\left(xy-yz\right)+\left(z-x\right)\left(xz-yz\right)\)

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

\(=\left(xy-yz\right)\left(-y+z\right)\)

mơn bn nha ^^

nh sáng nay lên lp thầy chữa bài thì kq nó k như z, cả cách lm nx :v

kq là: ( z - y )( x - z)( y - x )

[ вơ đắйǥ ] вé เςë ⁀ᶜᵘᵗᵉ

Ukm cảm ơn nhé quên mất đoạn cuối vẫn phân tích đc nữa

xy(x-y)+yz(y-z)+xz(x-z)

=y.[x.(x-y)+z.(y-z)]+xz(x-z)

=y.(x²-xy+zy-z²)+xz.(x-z)

=y.[(x²-z²)+(-xy+zy)]+xz.(x-z)

=y.[(x-z)(x+z)-y.(x-z)]+xz.(x-z)

=y.(x-z)(x+z-y)+xz.(x-z)

=(x-z)[y.(x+z-y)+xz]

=(x-z)(xy+yz-y²+xz)

xy(x+y)+yz(y+z)+xz(x+z)+2xyz 
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 
= xy(x + y) + yz(y + z + x) + xz(x + z + y) 
= xy(x + y) + z(x + y + z)(y + x) 
= (x + y)(xy + zx + zy + z²) 
= (x + y)[x(y + z) + z(y + z)] 
= (x + y)(y + z)(z + x)

xy(x+y)+yz(y+z)+xz(x+z)+2xyz 
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

tao có \(xz\left(z-x\right)+yz\left(y+z\right)-xy\left(x+y\right)=xz\left(z-x\right)+yz\left(y+x+z-x\right)-xy\left(x+y\right)=xz\left(z-x\right)+yz\left(z-x\right)+yz\left(x+y\right)-xy\left(x+y\right)\)

\(\left(z-x\right)\left(xz+yz\right)+\left(x+y\right)\left(yz-xy\right)=\left(z-x\right)z\left(x+y\right)+\left(x+y\right)y\left(z-x\right)=\left(z-x\right)\left(x+y\right)\left(z+y\right)\)

nếu mình giải khó hiểu thì cho mình xin lỗi nhé

\(xz\left(z-x\right)+yz\left(y+z\right)-xy+\left(x+y\right)\)

\(=xz^2-x^2z+yx\left(y+z\right)-xy\left(x+y\right)\)

\(=xz^2-x^2z+zy^2+z^2y-xy\left(x+y\right)\)

\(=xz^2-x^2z+zy^2+z^2y-x^2y-xy^2\)

P/s: ko chắc

Wrecking Ball giải chưa đúng đâu nha 

Nhưng đây là bài lớp 8, bạn mới học lớp 7 nên mình k động viên thui nha ( mình ko có điểm hỏi đáp đâu )

nhu the nay:

   ( xy( x + y )+ xyz )+( yz( y + z )+ xyz )+( xz( a +c )+ xyz)

= xy( x+y+z )+ yz( x + y + z )+ xz( x + y + z )

= ( x + y + z)( xy + yz +zx )

xong rui do dung thi ****.

a) \(xy+y-2x-2\)

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

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

b) \(xy+1+x+y\)

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

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

c) \(x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)

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

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)