\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\\left(\sqrt{2}-1\right)x+y=\sqrt{2}-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\sqrt{2}-\sqrt{2}+1\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1}{2}\\x+\left(\sqrt{2}+1\right)\dfrac{-1}{2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1}{2}\\x-\dfrac{\sqrt{2}+1}{2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1}{2}\\x=\dfrac{\sqrt{2}+3}{2}\end{matrix}\right.\)
Vậy S=\(\left\{\left(\dfrac{\sqrt{2}+3}{2};\dfrac{-1}{2}\right)\right\}\)
