\(\left(x^3+3x^2y+3xy^2+y^3-z^3\right):\left(x+y-z\right)\\ =\left[\left(x+y\right)^3-z^3\right]:\left(x+y-z\right)\\ =\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]:\left(x+y-z\right)\\ =x^2+2xy+y^2+xz+yz+z^2\)

Vậy chọn A 

Lấy 3 lần pt dưới cộng pt trên ta được :
\(4x^2+4y^2+z^2+2yz-4xz-4xy=0\)

\(\Leftrightarrow\left(2x-y-z\right)^2+3y^2=0\)

\(\Leftrightarrow\hept{\begin{cases}y=0\\2x-y-z=0\end{cases}\Rightarrow\hept{\begin{cases}y=0\\z=2x\end{cases}}}\)

\(\Rightarrow x^2+4x^2-2x^2=3\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=1;z=2\\x=-1;z=-2\end{cases}}\)

Lấy 3 lần pt dưới cộng pt trên ta được:

\(4x^2+4y^2+z^2+2yz-4xz-4xy=0\)

\(\Leftrightarrow\left(2x-y-z\right)^2+3y^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-y-z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=0\\z=2x\end{matrix}\right.\)

\(\Rightarrow x^2+4x^2-2x^2=3\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1;z=2\\x=-1;z=-2\end{matrix}\right.\)

xz-yz-x²+2xy-y²

=(xz-yz)-(x²-2xy+y²)

=z(x-y)-(x-y)²

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

đề ???

bí à bạn

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

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

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

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

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

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

\(x^3-3x^2+3x-9=x^2\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x^2+3\right)\)

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

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

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

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

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

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