1/ \(\left\{{}\begin{matrix}x\sqrt{2}-y\sqrt{3}=1\\x+y\sqrt{3}=\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(\sqrt{2}+1\right)=1+\sqrt{2}\\x+y\sqrt{3}=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{\sqrt{2}-1}{\sqrt{3}}\end{matrix}\right.\)
vậy hệ phương trình có ngiệm (x;y)=(1;\(\frac{\sqrt{2}-1}{\sqrt{3}}\))
2/ \(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\\left(\sqrt{2}-1\right)x+y=\sqrt{2}-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=-1\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-1}{2}\\x=\frac{3+\sqrt{2}}{2}\end{matrix}\right.\)
vậy hệ phương trình có nghiệm (x;y)=\(\left(\frac{3+\sqrt{2}}{2};\frac{-1}{2}\right)\)
