Nhận thấy \(x=y=0\) là 1 nghiệm
Với \(xy\ne0\) hệ tương đương:
\(\left\{{}\begin{matrix}\frac{1}{x^2}+\frac{1}{y^2}=2\\\left(\frac{x+y}{xy}\right)\left(\frac{1+xy}{xy}\right)=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{1}{x}+\frac{1}{y}\right)^2-\frac{2}{xy}=2\\\left(\frac{1}{x}+\frac{1}{y}\right)\left(1+\frac{1}{xy}\right)=4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\frac{1}{x}+\frac{1}{y}\\b=\frac{1}{xy}\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-2b=2\\a\left(b+1\right)=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-2\left(b+1\right)=0\\b+1=\frac{4}{a}\end{matrix}\right.\)
\(\Rightarrow a^2-\frac{8}{a}=0\Leftrightarrow a=3\Rightarrow b=\frac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=3\\\frac{1}{xy}=\frac{1}{3}\end{matrix}\right.\) bạn tự giải nốt
