\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=2xy+4\\x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=4\\x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y=2\\x+y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x-y=-2\\x+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=2\left(xy+2\right)\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-2xy+y^2=4\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-y\right)^2=4\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=2\\x-y=-2\end{matrix}\right.\\x+y=6\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y=2\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x-y=-2\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy (x;y)={(2;4);(4;2)}
