\(\left\{{}\begin{matrix}\sqrt{2}x-3y=11\\\left(1-\sqrt{2}\right)x+\left(1+\sqrt{2}\right)y=-4\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}\left(1-\sqrt{2}\right)x-3\left(1-\sqrt{2}\right)y=11-11\sqrt{2}\\\sqrt{2}\left(1-\sqrt{2}\right)x+\sqrt{2}\left(1+\sqrt{2}\right)y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(-3+3\sqrt{2}-\sqrt{2}-2\right)y=19-11\sqrt{2}\\\sqrt{2}\left(1-\sqrt{2}\right)x+\sqrt{2}\left(1+\sqrt{2}\right)y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{19-11\sqrt{2}}{2\sqrt{2}-5}=-3+\sqrt{2}\\\left(\sqrt{2}-2\right)x+\left(\sqrt{2}+2\right)\left(-3+\sqrt{2}\right)=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3+\sqrt{2}\\\left(\sqrt{2}-2\right)x=-12-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3+\sqrt{2}\\x=\dfrac{-4+\sqrt{2}}{\sqrt{2}-2}=3+\sqrt{2}\end{matrix}\right.\)
Vậy hpt có nghiệm duy nhất là:
(x;y) = (\(3+\sqrt{2};-3+\sqrt{2}\))
