ta có : \(\left\{{}\begin{matrix}x+2y=5\\x^2+2y^2-2xy=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\\left(5-2y\right)^2+2y^2-2\left(5-2y\right)y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\4y^2-20y+25+2y^2-10y+4y^2-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\10x^2-30y+20=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\\left[{}\begin{matrix}y=2\\y=1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}y=2\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\x=3\end{matrix}\right.\end{matrix}\right.\)
vậy phương trình có 2 nghiệm \(\left(1;2\right)\) và \(\left(3;1\right)\)
