Đáp án:
\(\left( {x;y} \right) = \left( {\dfrac{1}{6}; - \dfrac{5}{{12}}} \right)\)
Giải thích các bước giải:
Đặt \(\left\{ \begin{array}{l}\dfrac{1}{{x - 2y}} = a\\\dfrac{1}{{x + 2y}} = b\end{array} \right.\) ta được \(\left\{ \begin{array}{l}6a + 2b = 3\\5a + 4b = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = - \dfrac{3}{2}\end{array} \right.\)
Suy ra \(\left\{ \begin{array}{l}\dfrac{1}{{x - 2y}} = 1\\\dfrac{1}{{x + 2y}} = - \dfrac{3}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2y = 1\\x + 2y = - \dfrac{2}{3}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{1}{6}\\y = - \dfrac{5}{{12}}\end{array} \right.\)
Vậy hệ có nghiệm \(\left( {x;y} \right) = \left( {\dfrac{1}{6}; - \dfrac{5}{{12}}} \right)\)
