\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)x-\sqrt{2}\left(1\right)\\x+\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}-1\right)x-\sqrt{2}\right]=1\left(2\right)\end{matrix}\right.\)
Xét phương trình (2) ta có :
\(x+\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}-1\right)x-\sqrt{2}\right]=1\)
\(\Leftrightarrow x+\left(\sqrt{2}+1\right)\left(x\sqrt{2}-x-\sqrt{2}\right)=1\)
\(\Leftrightarrow x+2x-x\sqrt{2}-2+x\sqrt{2}-x-\sqrt{2}=1\)
\(\Leftrightarrow2x-2-\sqrt{2}=1\)
\(\Leftrightarrow2x=3+\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{3+\sqrt{2}}{2}\)
Thay \(x=\dfrac{3+\sqrt{2}}{2}\) vào phương trình (1) ta có :
\(y=\left(\sqrt{2}-1\right)\left(\dfrac{3+\sqrt{2}}{2}\right)-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+\sqrt{2}\right)}{2}-\dfrac{2\sqrt{2}}{2}\)
\(=\dfrac{3\sqrt{2}-1-\sqrt{2}-2\sqrt{2}}{2}\)
\(=-\dfrac{1}{2}\)
Vậy \(x=\dfrac{3+\sqrt{2}}{2}\) và \(y=-\dfrac{1}{2}\)
