Đáp án: $(x,y)=(1;\dfrac{-1}{2})$
Giải thích các bước giải:
ĐK: $x\ne 2y$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
x + \dfrac{x}{{x - 2y}} = \dfrac{3}{2}\\
- y + \dfrac{y}{{x - 2y}} = \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{x}{{x - 2y}} = \dfrac{3}{2}\\
- 2y + \dfrac{{2y}}{{x - 2y}} = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{x}{{x - 2y}} = \dfrac{3}{2}\\
x + \dfrac{x}{{x - 2y}} - \left( { - 2y + \dfrac{{2y}}{{x - 2y}}} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{x}{{x - 2y}} = \dfrac{3}{2}\\
x + 2y = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 2y\\
- 2y + \dfrac{{ - 2y}}{{ - 2y - 2y}} = \dfrac{3}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 2y\\
- 2y + \dfrac{1}{2} = \dfrac{3}{2}\left( {y \ne 0} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 2y\\
y = \dfrac{{ - 1}}{2}(tm)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = \dfrac{{ - 1}}{2}
\end{array} \right.\left( {tm} \right)
\end{array}$
Vậy hệ có nghiệm duy nhất là: $(x,y)=(1;\dfrac{-1}{2})$
