\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2=26\\3x^2-2y^2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2+3x^2-2y^2=20\\x^2+y^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x^2=20\\x^2+y^2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=4\\y^2=9\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=\pm2\\y=\pm3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=13\\3x^2-2y^2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2=26\\3x^2-2y^2=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x^2=20\\x^2+y^2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;3\right)\)
\(\left\{{}\begin{matrix}x^2+y^2=13\\3x^2-2y^2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2=26\\3x^2-2y^2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x^2=20\\3x^2-2y^2=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=-3\\y=3\\y=-3\\y=3\end{matrix}\right.\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là \(\left(-2;-3\right),\left(-2;3\right),\left(2;-3\right),\left(2;3\right)\)