Đáp án:

 b) \(\left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne 3;y \ne  - 1\\
\left\{ \begin{array}{l}
\dfrac{{20}}{{x - 3}} + \dfrac{{25}}{{y + 1}} = 10\\
\dfrac{{ - 20}}{{x - 3}} - \dfrac{4}{{y + 1}} =  - \dfrac{{29}}{5}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{25 - 4}}{{y + 1}} = \dfrac{{21}}{5}\\
\dfrac{{20}}{{x - 3}} + \dfrac{{25}}{{y + 1}} = 10
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y + 1 = 5\\
\dfrac{{20}}{{x - 3}} + \dfrac{{25}}{{y + 1}} = 10
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4\\
x = 7
\end{array} \right.\\
c)DK:x;y \ne 0\\
\left\{ \begin{array}{l}
 - \dfrac{4}{x} - \dfrac{6}{y} =  - 24\\
\dfrac{{15}}{x} + \dfrac{6}{y} = 57
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{15 - 4}}{x} = 33\\
\dfrac{{15}}{x} + \dfrac{6}{y} = 57
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{1}{3}\\
y = \dfrac{1}{2}
\end{array} \right.\\
b)DK:x;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{ - 2}}{x} - \dfrac{2}{y} =  - \dfrac{5}{3}\\
\dfrac{3}{x} + \dfrac{2}{y} = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{3 - 2}}{x} = \dfrac{1}{3}\\
\dfrac{3}{x} + \dfrac{2}{y} = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\\
d)DK:x \ne \dfrac{{3y}}{2};x \ne  - y\\
\left\{ \begin{array}{l}
\dfrac{2}{{2x - 3y}} + \dfrac{3}{{x + y}} = 4\\
 - \dfrac{2}{{2x - 3y}} + \dfrac{4}{{x + y}} =  - 18
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{7}{{x + y}} =  - 14\\
\dfrac{2}{{2x - 3y}} + \dfrac{3}{{x + y}} = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x + y =  - \dfrac{1}{2}\\
\dfrac{2}{{2x - 3y}} + \dfrac{3}{{x + y}} = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x =  - \dfrac{1}{2} - y\\
\dfrac{2}{{2\left( { - \dfrac{1}{2} - y} \right) - 3y}} + \dfrac{3}{{ - \dfrac{1}{2} - y + y}} = 4\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to \dfrac{2}{{ - 1 - 5y}} + \dfrac{3}{{ - \dfrac{1}{2}}} = 4\\
 \to y =  - \dfrac{6}{{25}}\\
 \to x =  - \dfrac{{13}}{{50}}
\end{array}\)