Đáp án:
2) \(\left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne - y;y \ne 1\\
Đặt:\left\{ \begin{array}{l}
\dfrac{1}{{x + y}} = a\\
\dfrac{1}{{y - 1}} = b
\end{array} \right.\\
Hpt \to \left\{ \begin{array}{l}
4a + b = 5\\
a - 2b = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8a + 2b = 10\\
a - 2b = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8a + a = 10 - 1\\
a - 2b = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
9a = 9\\
a - 2b = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = 1\\
b = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{{x + y}} = 1\\
\dfrac{1}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + y = 1\\
y - 1 = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 2\\
x = - 1
\end{array} \right.\\
2)DK:x \ne 0\\
Đặt:\left\{ \begin{array}{l}
\dfrac{1}{x} = a\\
y = b
\end{array} \right.\\
Hpt \to \left\{ \begin{array}{l}
\dfrac{3}{2}a - b = 6\\
a + 2b = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
b = \dfrac{3}{2}a - 6\\
a + 3a - 12 = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4a = 8\\
b = \dfrac{3}{2}a - 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = 2\\
b = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{x} = 2\\
y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = - 3
\end{array} \right.
\end{array}\)
