Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne \dfrac{3}{2};x \ne 0\\
\dfrac{1}{{2x - 3}} - \dfrac{3}{{x\left( {2x - 3} \right)}} = \dfrac{5}{x}\\
 \Rightarrow \dfrac{{x - 3}}{{x\left( {2x - 3} \right)}} = \dfrac{{5\left( {2x - 3} \right)}}{{x\left( {2x - 3} \right)}}\\
 \Rightarrow x - 3 = 10x - 15\\
 \Rightarrow 10x - x = 15 - 3\\
 \Rightarrow 9x = 12\\
 \Rightarrow x = \dfrac{4}{3}\left( {tmdk} \right)\\
b)Dkxd:x \ne 2;x \ne 0\\
\dfrac{{x + 2}}{{x - 2}} - \dfrac{1}{x} = \dfrac{2}{{x\left( {x - 2} \right)}}\\
 \Rightarrow \dfrac{{x\left( {x + 2} \right) - \left( {x - 2} \right)}}{{x\left( {x - 2} \right)}} = \dfrac{2}{{x\left( {x - 2} \right)}}\\
 \Rightarrow {x^2} + 2x - x + 2 = 2\\
 \Rightarrow {x^2} + x = 0\\
 \Rightarrow x\left( {x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x =  - 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - 1\\
c)Dkxd:x \ne 2;x \ne  - 2\\
\dfrac{{x + 1}}{{x - 2}} + \dfrac{{x - 1}}{{x + 2}} = \dfrac{{2\left( {{x^2} + 2} \right)}}{{{x^2} - 4}}\\
 \Rightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) + \left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2{x^2} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 \Rightarrow {x^2} + 3x + 2 + {x^2} - 3x + 2 = 2{x^2} + 4\\
 \Rightarrow 4 = 4\left( {luôn\,đúng} \right)\\
Vậy\,pt\,đúng\,\forall x \ne  - 2;x \ne 2\\
d)Dkxd:x \ne \dfrac{2}{7}\\
\left( {2x + 3} \right).\left( {\dfrac{{3x + 8}}{{2 - 7x}} + 1} \right) = \left( {x - 5} \right)\left( {\dfrac{{3x + 8}}{{2 - 7x}} + 1} \right)\\
 \Rightarrow \left( {2x + 3} \right).\dfrac{{3x + 8 + 2 - 7x}}{{2 - 7x}} = \left( {x - 5} \right).\dfrac{{3x + 8 + 2 - 7x}}{{2 - 7x}}\\
 \Rightarrow \left( {2x + 3} \right).\left( {10 - 4x} \right) - \left( {x - 5} \right).\left( {10 - 4x} \right) = 0\\
 \Rightarrow \left( {10 - 4x} \right).\left( {2x + 3 - x + 5} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{10}}{4} = \dfrac{5}{2}\left( {tm} \right)\\
x =  - 8\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - 8/x = \dfrac{5}{2}
\end{array}$