Giải hệ phương trình: 8(x^2 + y^2) + 4xy + 5/(x + y)^2 = 13; 2y + 1/(x + y) = 1
Câu hỏi:
Giải hệ phương trình: \(\left\{ \begin{array}{l}8\left( {{x^2} + {y^2}} \right) + 4xy + \frac{5}{{{{\left( {x + y} \right)}^2}}} = 13\\2y + \frac{1}{{x + y}} = 1\end{array} \right.\).
Trả lời:
Đặt x + y = a; x – y = b
Ta có hệ phương trình:
\(\left\{ \begin{array}{l}4\left( {{a^2} + {b^2}} \right) + {a^2} - {b^2} + \frac{5}{{{a^2}}} = 13\\a + b + \frac{1}{a} = 1\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}5{\left( {a + \frac{1}{a}} \right)^2} + 3{b^2} = 23\\a + \frac{1}{a} = 1 - b\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}5{\left( {1 - b} \right)^2} + 3{b^2} = 23\\a + \frac{1}{a} = 1 - b\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}8{b^2} - 10b - 18 = 0\\a + \frac{1}{a} = 1 - b\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}\left[ \begin{array}{l}b = - 1\\b = \frac{9}{4}\end{array} \right.\\a + \frac{1}{a} = 1 - b\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\left\{ \begin{array}{l}b = - 1\\a + \frac{1}{a} = 2\end{array} \right.\\\left\{ \begin{array}{l}b = \frac{9}{4}\\a + \frac{1}{a} = \frac{{ - 5}}{4}\end{array} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}b = - 1\\a = - 1\end{array} \right.\\\left\{ \begin{array}{l}b = \frac{9}{4}\\{a^2} + \frac{5}{4}a + 1 = 0\end{array} \right.\end{array} \right.\)
Vậy a = b = –1.
