Đáp án:
\(\left\{ \begin{array}{l}
x = \dfrac{1}{4}\\
y = - \dfrac{{11}}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 1;y \ne - 2\\
Đặt:\left\{ \begin{array}{l}
\dfrac{1}{{x - 1}} = a\\
\dfrac{1}{{y + 2}} = b
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{a} + 1\\
y = \dfrac{1}{b} - 2
\end{array} \right.\\
Hpt \to \left\{ \begin{array}{l}
\left( {\dfrac{1}{a} + 1 + 1} \right)a + 3.\left( {\dfrac{1}{b} - 2} \right).b = 7\\
2a - 5b = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a + 2a + 3 - 6b = 7\\
2a - 5b = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3a - 6b = 4\\
2a - 5b = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = - \dfrac{4}{3}\\
b = - \dfrac{4}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{{x - 1}} = - \dfrac{4}{3}\\
\dfrac{1}{{y + 2}} = - \dfrac{4}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{4}\\
y = - \dfrac{{11}}{4}
\end{array} \right.
\end{array}\)
