Giải hệ phương trình:

\[\left\{ \begin{array}{l}xy + x + y = 11\\{x^2}y + x{y^2} = 30\end{array} \right.\]

Lời giải

\[\left\{ \begin{array}{l}xy + x + y = 11\\{x^2}y + x{y^2} = 30\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}xy + \left( {x + y} \right) = 11\\xy\left( {x + y} \right) = 30\end{array} \right.\] (*)

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

Hệ phương trình (*) trở thành

\[\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}a + b = 11\\ab = 30\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\a\left( {11 - a} \right) = 30\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\{a^2} - 11a + 30 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 11 - a\\\left( {a - 5} \right)\left( {a - 6} \right) = 0\end{array} \right.\]

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

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

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

\[ \Leftrightarrow \left\{ \begin{array}{l}y = 6 - x\\{x^2} - 6x + 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 6 - x\\\left( {x - 1} \right)\left( {x - 5} \right) = 0\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}y = 6 - x\\\left[ \begin{array}{l}x = 1\\x = 5\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 1\\y = 5\end{array} \right.\\\left\{ \begin{array}{l}x = 5\\y = 1\end{array} \right.\end{array} \right.\].

Vậy cặp nghiệm (x; y) của hệ phương trình là: \[\left( {x;\;y} \right) = \left\{ {\left( {2;\;3} \right),\;\left( {3;\;2} \right),\;\left( {1;\;5} \right),\;\left( {5;\;1} \right)} \right\}\].