Hệ phương trình $\left\{ {\begin{array}{*{20}{c}}{x.y + x + y = 11}\\{{x^2}y + x{y^2} = 30}\end{array}} \right.$

Ta có $\left\{ {\begin{array}{*{20}{c}}{x.y + x + y = 11}\\{{x^2}y + x{y^2} = 30}\end{array}} \right. \Leftrightarrow \left\{ \begin{array}{l}xy + x + y = 11\\xy\left( {x + y} \right) = 30\end{array} \right.$

Đặt \(S = x + y;P = xy\,\left( {{S^2} \ge 4P} \right)\) ta có hệ \(\left\{ \begin{array}{l}S + P = 11\\S.P = 30\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}S = 11 - P\\\left( {11 - P} \right).P = 30\,\,\,\left( 1 \right)\end{array} \right.\)

Xét phương trình \(\left( 1 \right):\)

 \(\,11P - {P^2} - 30 = 0 \Leftrightarrow {P^2} - 11P + 30 = 0 \Leftrightarrow \left( {P - 5} \right)\left( {P - 6} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}P = 5 \Rightarrow S = 6\\P = 6 \Rightarrow S = 5\end{array} \right.\) ( tm \({S^2} \ge 4P\))

Với \(P = 5;S = 6 \Rightarrow \left\{ \begin{array}{l}xy = 5\\x + y = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 6 - x\\x\left( {6 - x} \right) - 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 6 - x\\{x^2} - 6x + 5 = 0\end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 1\\y = 5\end{array} \right.\\\left\{ \begin{array}{l}x = 5\\y = 1\end{array} \right.\end{array} \right.\)

Với \(P = 6;S = 5\) \( \Rightarrow \left\{ \begin{array}{l}xy = 6\\x + y = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\x\left( {5 - x} \right) - 6 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\{x^2} - 5x + 6 = 0\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\\\left\{ \begin{array}{l}x = 3\\y = 2\end{array} \right.\end{array} \right.\)

Vậy hệ phương trình có bốn nghiệm $\left( {2;3} \right),\left( {3;2} \right),\left( {1;5} \right),\left( {5;1} \right).$