Giải phương trình sau: (phương trình đối xứng loại 2) $\left \{ {{2x+ \frac{1}{y} = \frac{3}{x}} \atop {2y+ \frac{1}{x} = \frac{3}{y} }} \right.$

Đáp án:

$\begin{array}{l}
\left\{ \begin{array}{l}
2x + \frac{1}{y} = \frac{3}{x}\\
2y + \frac{1}{x} = \frac{3}{y}
\end{array} \right.\left( {x,y \ne 0} \right)\\
 \Rightarrow 2x + \frac{1}{y} - 2y - \frac{1}{x} = \frac{3}{x} - \frac{3}{y}\\
 \Rightarrow 2\left( {x - y} \right) + \left( {\frac{1}{y} - \frac{1}{x}} \right) = 3\left( {\frac{1}{x} - \frac{1}{y}} \right)\\
 \Rightarrow 2\left( {x - y} \right) - \frac{{x - y}}{{xy}} + 3.\frac{{x - y}}{{xy}} = 0\\
 \Rightarrow \left( {x - y} \right)\left( {2 - \frac{1}{{xy}} + \frac{3}{{xy}}} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = y\\
2 + \frac{2}{{xy}} = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = y\left( 1 \right)\\
xy =  - 1 \Rightarrow  - x = \frac{1}{y}\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Rightarrow 2x + \frac{1}{x} = \frac{3}{x} \Rightarrow 2x - \frac{2}{x} = 0 \Rightarrow x = y =  \pm 1\\
\left( 2 \right) \Rightarrow 2x - x = \frac{3}{x} \Rightarrow x = \frac{3}{x} \Rightarrow x =  \pm \sqrt 3  \Rightarrow y = \frac{{ \mp 1}}{{\sqrt 3 }}
\end{array}$