Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
\dfrac{3}{{x + y}} - \dfrac{1}{{x - y}} = 5\\
\dfrac{2}{{x + y}} + \dfrac{3}{{x - y}} = 18
\end{array} \right.\\
Đặt:\dfrac{1}{{x + y}} = a;\dfrac{1}{{x - y}} = b\\
\Rightarrow \left\{ \begin{array}{l}
3.a - b = 5\\
2.a + 3.b = 18
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
9a - 3b = 15\\
2.a + 3b = 18
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
11.a = 33\\
b = 3a - 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 3\\
b = 3.3 - 5 = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + y}} = 3\\
\dfrac{1}{{x - y}} = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + y = \dfrac{1}{3}\\
x - y = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2.x = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{{12}}\\
y = x - \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{7}{{24}}\\
y = \dfrac{7}{{24}} - \dfrac{1}{4} = \dfrac{1}{{24}}
\end{array} \right.\\
Vay\,x = \dfrac{7}{{24}};y = \dfrac{1}{{24}}\\
b)\left\{ \begin{array}{l}
\dfrac{3}{{x + 2}} + \dfrac{1}{y} = 5\\
\dfrac{1}{{x + 2}} - \dfrac{3}{y} = 1
\end{array} \right.\\
Đặt:\left\{ \begin{array}{l}
\dfrac{1}{{x + 2}} = a\\
\dfrac{1}{y} = b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3.a + b = 5\\
a - 3.b = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3a + b = 5\\
3.a - 9b = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
10.b = 2\\
a = 3b + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = \dfrac{1}{5}\\
a = \dfrac{8}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + 2}} = a = \dfrac{8}{5}\\
\dfrac{1}{y} = b = \dfrac{1}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + 2 = \dfrac{5}{8}\\
y = 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - 11}}{8}\\
y = 5
\end{array} \right.\\
Vay\,x = - \dfrac{{11}}{8};y = 5
\end{array}$
