Đáp án:
\(\left( {x;y} \right) \in \left\{ {\left( {1;1} \right),\left( {2; - \dfrac{3}{2}} \right)} \right\}\)
Giải thích các bước giải:
Ta có: \(\left\{ \begin{array}{l}x\left( {x + y + 1} \right) - 3 = 0\\{\left( {x + y} \right)^2} - \dfrac{5}{{{x^2}}} + 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x\left( {x + y} \right) + x - 3 = 0\\{x^2}{\left( {x + y} \right)^2} + {x^2} - 5 = 0\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}a = x\left( {x + y} \right)\\b = x\end{array} \right.\) ta được
\(\begin{array}{l}\left\{ \begin{array}{l}a + b - 3 = 0\\{a^2} + {b^2} - 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 3 - a\\{a^2} + {\left( {3 - a} \right)^2} - 5 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3 - a\\{a^2} + {a^2} - 6a + 9 - 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 3 - a\\2{a^2} - 6a + 4 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3 - a\\\left[ \begin{array}{l}a = 1\\a = 2\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}a = 1,b = 2\\a = 2,b = 1\end{array} \right.\end{array}\)
TH1: \(\left\{ \begin{array}{l}a = 1\\b = 2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x\left( {x + y} \right) = 1\\x = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = - \dfrac{3}{2}\end{array} \right.\)
TH2: \(\left\{ \begin{array}{l}a = 2\\b = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x\left( {x + y} \right) = 2\\x = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\)
Vậy hệ có nghiệm \(\left( {x;y} \right) \in \left\{ {\left( {1;1} \right),\left( {2; - \dfrac{3}{2}} \right)} \right\}\)
