Giải:
\(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow\left(2x-2y\right)-\left(x^2-2xy+y^2\right)=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left[2-\left(x-y\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\)
Vậy ...
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\(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-y\right)=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x=2-y\end{matrix}\right.\)
Vậy \(x=y\) hoặc \(x=2-y\)
