ĐKXĐ: \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+4=0\\x\left(y+\dfrac{1}{y}\right)+\dfrac{1}{x}\left(y+\dfrac{1}{y}\right)-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=-4\\\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=u\\y+\dfrac{1}{y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=-4\\uv=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}v=-4-u\\uv=4\end{matrix}\right.\)
\(\Leftrightarrow u\left(-4-u\right)=4\Leftrightarrow u^2+4u+4=0\)
\(\Leftrightarrow\left(u+2\right)^2=0\)
\(\Rightarrow u=-2\Rightarrow v=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=-2\\y+\dfrac{1}{y}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+2x+1=0\\y^2+2y+1=0\end{matrix}\right.\)
\(\Rightarrow x=y=-1\)
