Hệ phương trình \(\left\{ \begin{array}{l}x + y + 2xy =  - 8\\{x^2} + {y^2} = 10\end{array} \right.\)có bao nhiêu nghiệm?

+ Ta có \(\left\{ \begin{array}{l}x + y + 2xy =  - 8\\{x^2} + {y^2} = 10\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y + 2xy =  - 8\\{\left( {x + y} \right)^2} - 2xy = 10\end{array} \right.\)

+ Đặt \(S = x + y;P = xy\) ta được hệ phương trình \(\left\{ \begin{array}{l}S + 2P =  - 8\\{S^2} - 2P = 10\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}S + 2P =  - 8\\{S^2} + S - 2 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}S + 2P =  - 8\\\left( {S - 1} \right)\left( {S + 2} \right) = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}P = \dfrac{{ - 8 - S}}{2}\\\left[ \begin{array}{l}S =  - 2\\S = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}S =  - 2\\P =  - 3\end{array} \right.\\\left\{ \begin{array}{l}S = 1\\P =  - \dfrac{9}{2}\end{array} \right.\end{array} \right.\)  (tm \({S^2} \ge 4P\))

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

+)  Với \(\left\{ \begin{array}{l}S = 1\\P =  - \dfrac{9}{2}\end{array} \right.\) thì  \(\left\{ \begin{array}{l}xy =  - \dfrac{9}{2}\\x + y = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 1 - x\\x\left( {1 - x} \right) =  - \dfrac{9}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 1 - x\\{x^2} - x - \dfrac{9}{2} = 0\left( * \right)\end{array} \right.\)

Nhận thấy phương trình \(\left( * \right)\) có \(\Delta  = 19 > 0\) nên có hai nghiệm \(\left[ \begin{array}{l}x = \dfrac{{1 + \sqrt {19} }}{2} \Rightarrow y = \dfrac{{1 - \sqrt {19} }}{2}\\x = \dfrac{{1 - \sqrt {19} }}{2} \Rightarrow y = \dfrac{{1 + \sqrt {19} }}{2}\end{array} \right.\)

Vậy hệ phương trình có bốn nghiệm \(\left( {1; - 3} \right);\left( { - 3;1} \right);\left( {\dfrac{{1 + \sqrt {19} }}{2};\dfrac{{1 - \sqrt {19} }}{2}} \right);\left( {\dfrac{{1 - \sqrt {19} }}{2};\dfrac{{1 + \sqrt {19} }}{2}} \right)\)