\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)x-\left(\sqrt{2}+1\right)y=\left(\sqrt{2+1}\right)\sqrt{2}\\x+\left(\sqrt{2+1}\right)y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\left(\sqrt{2}+1\right)y=2+\sqrt{2}\left(1\right)\\x+\left(\sqrt{2}+1\right)y=1\left(2\right)\end{matrix}\right.\)
Cộng từng vế của (1) và (2) ta được: \(\Rightarrow2x=3+\sqrt{2}\Leftrightarrow x=\dfrac{3+\sqrt{2}}{2}\)
Thay vào (2) ta được: \(\Rightarrow\dfrac{3+\sqrt{2}}{2}+\left(\sqrt{2}+1\right)y=1\Leftrightarrow\left(\sqrt{2}+1\right)y=1-\dfrac{3+\sqrt{2}}{2}=\dfrac{-\sqrt{2}-1}{2}\)
\(\Leftrightarrow y=\dfrac{-\sqrt{2}-1}{2\left(\sqrt{2}+1\right)}=\dfrac{-1}{2}\) Vậy...
