Giải hệ phương trình:

\[\left\{ \begin{array}{l}x + y + xy = 11\\{x^2} + {y^2} + 3\left( {x + y} \right) = 28\end{array} \right.\]

Lời giải

\[\left\{ \begin{array}{l}x + y + xy = 11\\{x^2} + {y^2} + 3\left( {x + y} \right) = 28\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}\left( {x + y} \right) + xy = 11\\{\left( {x + y} \right)^2} - 2xy + 3\left( {x + y} \right) = 28\end{array} \right.\] (*)

Ta đặt: a = x + y và b = xy (Với a² ≥ − 4b)

Hệ phương trình (*) trở thành

\[\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}a + b = 11\\{a^2} - 2b + 3a = 28\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\{a^2} - 2\left( {11 - a} \right) + 3a = 28\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\{a^2} - 22 + 2a + 3a = 28\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\{a^2} + 5a - 50 = 0\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\\left( {a - 5} \right)\left( {a + 10} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\\left[ \begin{array}{l}a = 5\\a = - 10\end{array} \right.\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 5\\b = 11 - 5\end{array} \right.\\\left\{ \begin{array}{l}a = - 10\\b = 11 - \left( { - 10} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 5\\b = 6\end{array} \right.\\\left\{ \begin{array}{l}a = - 10\\b = 21\end{array} \right.\end{array} \right.\]

+ TH1: \[\left\{ \begin{array}{l}a = 5\\b = 6\end{array} \right.\]

\[ \Rightarrow \left\{ \begin{array}{l}x + y = 5\\xy = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\x\left( {5 - x} \right) = 6\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\{x^2} - 5x + 6 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\\left( {x - 2} \right)\left( {x - 3} \right) = 0\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}y = 5 - x\\\left[ \begin{array}{l}x = 2\\x = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\\\left\{ \begin{array}{l}x = 3\\y = 2\end{array} \right.\end{array} \right.\]

+ TH2: \[\left\{ \begin{array}{l}a = - 10\\b = 21\end{array} \right.\]

\[ \Rightarrow \left\{ \begin{array}{l}x + y = - 10\\xy = 21\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = - 10 - x\\x\left( { - 10 - x} \right) = 21\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}y = - 10 - x\\{x^2} + 10x + 21 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = - 10 - x\\\left( {x + 3} \right)\left( {x + 7} \right) = 0\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}y = - 10 - x\\\left[ \begin{array}{l}x = - 3\\x = - 7\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = - 3\\y = - 7\end{array} \right.\\\left\{ \begin{array}{l}x = - 7\\y = - 3\end{array} \right.\end{array} \right.\]

Vậy cặp nghiệm (x; y) của hệ phương trình là:

\[\left( {x;\;y} \right) = \left\{ {\left( {2;\;3} \right),\;\left( {3;\;2} \right),\;\left( { - 3;\; - 7} \right),\;\left( { - 7;\; - 3} \right)} \right\}\]