Giải hệ phương trình: \(\left\{ \begin{array}{l}x + y + \frac{1}{x} + \frac{1}{y} = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}\left( {x + \frac{1}{y}} \right) = xy + \frac{1}{{xy}}\end{array} \right.\)

Lời giải:

\(\left\{ \begin{array}{l}x + y + \frac{1}{x} + \frac{1}{y} = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}\left( {x + \frac{1}{y}} \right) = xy + \frac{1}{{xy}}\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left( {x + \frac{1}{y}} \right) + \left( {y + \frac{1}{x}} \right) = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}\left( {x + \frac{1}{y}} \right) = \left( {x + \frac{1}{y}} \right)\left( {y + \frac{1}{x}} \right)\end{array} \right.\)

Đặt \(x + \frac{1}{y} = a;y + \frac{1}{x} = b\)

Khi đó ta có hệ: \(\left\{ \begin{array}{l}a + b = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}a = ab\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\\frac{1}{4} + \frac{3}{2}\left( {\frac{9}{2} - b} \right) = \left( {\frac{9}{2} - b} \right)b\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\7 - 6b + {b^2} = 0\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\\left[ \begin{array}{l}b = 1\\b = - 7\end{array} \right.\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}a = \frac{7}{2}\\b = 1\end{array} \right.\\\left\{ \begin{array}{l}a = \frac{{23}}{2}\\b = - 7\end{array} \right.\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{7}{2}\\y + \frac{1}{x} = 1\end{array} \right.\\\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{{23}}{2}\\y + \frac{1}{x} = - 7\end{array} \right.\end{array} \right.\)

+ Với \[\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{7}{2}\\y + \frac{1}{x} = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{1}{{1 - \frac{1}{x}}} = \frac{7}{2}\\y = 1 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x\left( {1 - \frac{1}{x}} \right) + 1 = \frac{7}{2}\left( {1 - \frac{1}{x}} \right)\\y = 1 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{7}{2} - \frac{7}{{2x}}\\y = 1 - \frac{1}{x}\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}x + \frac{7}{{2x}} - \frac{7}{2} = 0\\y = 1 - \frac{1}{x}\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}2{x^2} + 7 - 2x = 0\left( {VN} \right)\\y = 1 - \frac{1}{x}\end{array} \right.\]

+ Với \(\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{{23}}{2}\\y + \frac{1}{x} = - 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{1}{{ - 7 - \frac{1}{x}}} = \frac{{23}}{2}\\y = - 7 - \frac{1}{x}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}7x = \frac{{161x}}{{2x}} + \frac{{23}}{{2x}}\\y = - 7 - \frac{1}{x}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}49{x^2} - 161x - 23 = 0\\y = - 7 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = \frac{{23 + 3\sqrt {69} }}{{14}}\\y = \frac{{ - 161 - 21\sqrt {69} }}{{46}}\end{array} \right.\\\left\{ \begin{array}{l}x = \frac{{23 - 3\sqrt {69} }}{{14}}\\y = \frac{{ - 161 + 21\sqrt {69} }}{{46}}\end{array} \right.\end{array} \right.\)