ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\) \(\Rightarrow a^2+a=6\Leftrightarrow a^2+a-6=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\Rightarrow b=1\\a=-3\Rightarrow b=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\frac{1}{y}=2\\\frac{x}{y}=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+\frac{1}{y}=-3\\\frac{x}{y}=6\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\frac{1}{y}=2\\x=y\end{matrix}\right.\\\left\{{}\begin{matrix}x+\frac{1}{y}=-3\\x=6y\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\frac{1}{x}=2\\x=y\end{matrix}\right.\\\left\{{}\begin{matrix}6y+\frac{1}{y}=-3\\x=6y\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-2x+1=0\\x=y\end{matrix}\right.\\\left\{{}\begin{matrix}6y^2+3y+1=0\\x=6y\left(vn\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x=y=1\)
