Giải hệ phương trình \(\left\{ \begin{array}{l}{x^2} + x + {y^2} + y = 18\\xy\left( {x + 1} \right)\left( {y + 1} \right) = 72\end{array} \right.\)

Lời giải:

\(\left\{ \begin{array}{l}{x^2} + x + {y^2} + y = 18\\xy\left( {x + 1} \right)\left( {y + 1} \right) = 72\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}x\left( {x + 1} \right) + y\left( {y + 1} \right) = 18\\x\left( {x + 1} \right).y\left( {y + 1} \right) = 72\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x\left( {x + 1} \right) = 12\\y\left( {y + 1} \right) = 6\end{array} \right.\\\left\{ \begin{array}{l}x\left( {x + 1} \right) = 6\\y\left( {y + 1} \right) = 12\end{array} \right.\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3;x = - 4\\y = 2;y = - 3\end{array} \right.\\\left\{ \begin{array}{l}x = 2;x = - 3\\y = 3;y = - 4\end{array} \right.\end{array} \right.\)

Vậy (x,y) = {(3,2);(3,−3);(−4;2);(−4;−3)} và hoán vị