Giải hệ phương trình \(\left\{ \begin{array}{l}y + x{y^2} = 6{{\rm{x}}^2}\\1 + {x^2}{y^2} = 5{{\rm{x}}^2}\end{array} \right.\).

Ta có: \(\left\{ \begin{array}{l}y + x{y^2} = 6{{\rm{x}}^2}\\1 + {x^2}{y^2} = 5{{\rm{x}}^2}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\frac{y}{{{x^2}}} + \frac{{{y^2}}}{x} = 6\\\frac{1}{{{x^2}}} + {y^2} = 5\end{array} \right.\)

Đặt \[{\rm{a}} = \frac{1}{x}\] ta có: \(\left\{ \begin{array}{l}{a^2}y + a{y^2} = 6\\{a^2} + {y^2} = 5\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}ay\left( {a + y} \right) = 6\\{\left( {a + y} \right)^2} - 2{\rm{a}}y = 5\end{array} \right.\)

Đặt ay = z, a + y = t

Ta có \(\left\{ \begin{array}{l}t{\rm{z}} = 6\\{t^2} - 2{\rm{z}} = 5\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{\rm{z}} = \frac{{{t^2} - 5}}{2}\\t.\frac{{{t^2} - 5}}{2} = 6\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{\rm{z}} = \frac{{{t^2} - 5}}{2}\\{t^3} - 5t - 12 = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{\rm{z}} = \frac{{{t^2} - 5}}{2}\\\left( {t - 3} \right)\left( {{t^2} + 3t + 4} \right) = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{\rm{z}} = \frac{{{t^2} - 5}}{2}\\t - 3 = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{\rm{z}} = 2\\t = 3\end{array} \right.\)

Suy ra \(\left\{ \begin{array}{l}ay = 2\\a + y = 3\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y\left( {3 - y} \right) = 2\\a = 3 - y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{y^2} - 3y + 2 = 0\\a = 3 - y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}y = 1\\y = 2\end{array} \right.\\a = 3 - y\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}y = 1\\a = 2\end{array} \right.\\\left\{ \begin{array}{l}y = 2\\a = 1\end{array} \right.\end{array} \right.\)

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

Vậy hệ phương trình có nghiệm (x; y) = (1; 2) hoặc \(\left( {x;y} \right) = \left( {\frac{1}{2};1} \right)\).