Hệ phương trình $\left\{ \begin{array}{l}{x^2}y + x{y^2} = 6\\xy + x + y = 5\end{array} \right.{\rm{  }}$

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

Đặt $S = x + y;P = xy\,\left( {{S^2} \ge 4P} \right)$ ta có hệ $\left\{ \begin{array}{l}S + P = 5\\S.P = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}S = 5 - P\\\left( {5 - P} \right).P = 6\,\,\left( 1 \right)\end{array} \right.$

Xét phương trình $\left( 1 \right):$

 $\,5P - {P^2} - 6 = 0 \Leftrightarrow {P^2} - 5P + 6 = 0 \Leftrightarrow \left( {P - 2} \right)\left( {P - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}P = 2 \Rightarrow S = 3\left( {tm\,{S^2} \ge 4P} \right)\\P = 3 \Rightarrow S = 2\left( {ktm\,{S^2} \ge 4P} \right)\end{array} \right.$

Với $P = 2;S = 3 \Rightarrow \left\{ \begin{array}{l}xy = 2\\x + y = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 3 - x\\x\left( {3 - x} \right) - 2 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 3 - x\\{x^2} - 3x + 2 = 0\end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 1\\y = 2\end{array} \right.\\\left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\end{array} \right.$

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