Giải hệ phương trình (căn2−1)x−y=căn2, x+(căn2+1)y=1

\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)x-\sqrt{2}\left(1\right)\\x+\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}-1\right)x-\sqrt{2}\right]=1\left(2\right)\end{matrix}\right.\)

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 \(y=-\dfrac{1}{2}\)