Đáp án:

$\begin{array}{l}
a)2,5\left( {x + \dfrac{2}{3}} \right) - \dfrac{{ - 15}}{{20}} = \left| {1 - 3\dfrac{1}{2}} \right|\\
 \Leftrightarrow \dfrac{5}{2}.\left( {x + \dfrac{2}{3}} \right) + \dfrac{3}{4} = \left| {1 - \dfrac{7}{2}} \right|\\
 \Leftrightarrow \dfrac{5}{2}x + \dfrac{5}{3} + \dfrac{3}{4} = \left| {\dfrac{{ - 5}}{2}} \right|\\
 \Leftrightarrow \dfrac{5}{2}x = \dfrac{5}{2} - \dfrac{5}{3} - \dfrac{3}{4}\\
 \Leftrightarrow \dfrac{5}{2}x = \dfrac{1}{{12}}\\
 \Leftrightarrow x = \dfrac{1}{{12}}.\dfrac{2}{5}\\
 \Leftrightarrow x = \dfrac{1}{{30}}\\
Vậy\,x = \dfrac{1}{{30}}\\
b)\dfrac{2}{3}:{\left( {1 - 2x} \right)^2} + {\left( {\dfrac{{ - 2}}{3}} \right)^2} = 3\dfrac{1}{9}\\
 \Leftrightarrow \dfrac{2}{3}:{\left( {2x - 1} \right)^2} + \dfrac{4}{9} = \dfrac{{28}}{9}\\
 \Leftrightarrow \dfrac{2}{3}:{\left( {2x - 1} \right)^2} = \dfrac{{28}}{9} - \dfrac{4}{9} = \dfrac{{24}}{9} = \dfrac{8}{3}\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} = \dfrac{2}{3}:\dfrac{8}{3}\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} = \dfrac{1}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = \dfrac{1}{2}\\
2x - 1 =  - \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{3}{2} \Leftrightarrow x = \dfrac{3}{4}\\
2x = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{3}{4};x = \dfrac{1}{4}\\
c)\left| {\dfrac{5}{2} - 2:x} \right| - \dfrac{{ - 3}}{4} = {\left( {\dfrac{3}{2}} \right)^2}\\
 \Leftrightarrow \left| {\dfrac{5}{2} - 2:x} \right| = \dfrac{9}{4} + \dfrac{{ - 3}}{4} = \dfrac{6}{4} = \dfrac{3}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{5}{2} - 2:x = \dfrac{3}{2}\\
\dfrac{5}{2} - 2:x = \dfrac{{ - 3}}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2:x = \dfrac{5}{2} - \dfrac{3}{2} = 1\\
2:x = \dfrac{5}{2} + \dfrac{3}{2} = 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = 2;x = \dfrac{1}{2}\\
d)3\left( {x - 2} \right) + 4\left( {\dfrac{5}{2} + x} \right) > 0\\
 \Leftrightarrow 3x - 6 + 10 + 4x > 0\\
 \Leftrightarrow 7x >  - 4\\
 \Leftrightarrow x >  - \dfrac{4}{7}\\
Vậy\,x > \dfrac{{ - 4}}{7}
\end{array}$