ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=4\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=8\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=4\\a^2+b^2=8\end{matrix}\right.\) \(\Rightarrow a^2+\left(4-a\right)^2-8=0\)
\(\Leftrightarrow2a^4-8a+8=0\Rightarrow a=2\Rightarrow b=2\)
\(\Rightarrow\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)