ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x+1}=u\\\dfrac{y}{y+1}=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}2u+v=\sqrt{2}\\u+3v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6u+3v=3\sqrt{2}\\u+3v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5u=3\sqrt{2}-1\\v=\sqrt{2}-2u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{3\sqrt{2}-1}{5}\\v=\dfrac{2-\sqrt{2}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{3\sqrt{2}-1}{5}\\\dfrac{y}{y+1}=\dfrac{2-\sqrt{2}}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=\left(3\sqrt{2}-1\right)x+3\sqrt{2}-1\\5y=\left(2-\sqrt{2}\right)y+2-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4+5\sqrt{2}}{6}\\y=\dfrac{8-5\sqrt{2}}{7}\end{matrix}\right.\)
