Đáp án:

$\begin{array}{l}
a){\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = \dfrac{1}{{81}}\\
 \Rightarrow {\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{3}} \right)^4} = {\left( { - \dfrac{1}{3}} \right)^4}\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{{2x}}{3} + \dfrac{1}{2} = \dfrac{1}{3}\\
\dfrac{{2x}}{3} + \dfrac{1}{2} =  - \dfrac{1}{3}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{{2x}}{3} = \dfrac{{ - 1}}{6}\\
\dfrac{{2x}}{3} =  - \dfrac{5}{6}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 1}}{4}\\
x =  - \dfrac{5}{4}
\end{array} \right.\\
b){\left( {\dfrac{3}{4} - \dfrac{{3x}}{5}} \right)^3} = \dfrac{8}{{125}}\\
 \Rightarrow \dfrac{3}{4} - \dfrac{{3x}}{5} = \dfrac{2}{5}\\
 \Rightarrow \dfrac{{3x}}{5} = \dfrac{3}{4} - \dfrac{2}{5} = \dfrac{7}{{20}}\\
 \Rightarrow x = \dfrac{7}{{20}}.\dfrac{5}{3}\\
 \Rightarrow x = \dfrac{7}{{12}}\\
Vay\,x = \dfrac{7}{{12}}\\
c)\dfrac{1}{3}:{\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{4}{3}\\
 \Rightarrow {\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{1}{3}:\dfrac{4}{3} = \dfrac{1}{4}\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{x}{2} + \dfrac{3}{4} = \dfrac{1}{2}\\
\dfrac{x}{2} + \dfrac{3}{4} =  - \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{x}{2} =  - \dfrac{1}{4}\\
\dfrac{x}{2} = \dfrac{{ - 5}}{4}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 1}}{2}\\
x =  - \dfrac{5}{2}
\end{array} \right.\\
d){\left( {\dfrac{5}{6} - \dfrac{{3x}}{2}} \right)^5} = \dfrac{{243}}{{1024}}\\
 \Rightarrow \dfrac{5}{6} - \dfrac{{3x}}{2} = \dfrac{3}{4}\\
 \Rightarrow \dfrac{{3x}}{2} = \dfrac{5}{6} - \dfrac{3}{4}\\
 \Rightarrow \dfrac{{3x}}{2} = \dfrac{1}{{12}}\\
 \Rightarrow x = \dfrac{1}{{18}}\\
Vay\,x = \dfrac{1}{{18}}
\end{array}$

$a){\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = \dfrac{1}{{81}}\\ ⇔ {\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{3}} \right)^4} = {\left( { - \dfrac{1}{3}} \right)^4}\\ ⇔ \left[ \begin{array}{l} \dfrac{{2x}}{3} + \dfrac{1}{2} = \dfrac{1}{3}\\ \dfrac{{2x}}{3} + \dfrac{1}{2} = - \dfrac{1}{3} \end{array} \right.\\ ⇔ \left[ \begin{array}{l} \dfrac{{2x}}{3} = \dfrac{{ - 1}}{6}\\ \dfrac{{2x}}{3} = - \dfrac{5}{6} \end{array} \right.\\ ⇔ \left[ \begin{array}{l} x = \dfrac{{ - 1}}{4}\\ x = - \dfrac{5}{4} \end{array} \right.\\ b){\left( {\dfrac{3}{4} - \dfrac{{3x}}{5}} \right)^3} = \dfrac{8}{{125}}\\ ⇔ \dfrac{3}{4} - \dfrac{{3x}}{5} = \dfrac{2}{5}\\ ⇔ \dfrac{{3x}}{5} = \dfrac{3}{4} - \dfrac{2}{5} = \dfrac{7}{{20}}\\ ⇔ x = \dfrac{7}{{20}}.\dfrac{5}{3}\\ ⇔ x = \dfrac{7}{{12}}\\ c)\dfrac{1}{3}:{\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{4}{3}\\ ⇔ {\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{1}{3}:\dfrac{4}{3} = \dfrac{1}{4}\\ ⇔ \left[ \begin{array}{l} \dfrac{x}{2} + \dfrac{3}{4} = \dfrac{1}{2}\\ \dfrac{x}{2} + \dfrac{3}{4} = - \dfrac{1}{2} \end{array} \right.\\ ⇔ \left[ \begin{array}{l} \dfrac{x}{2} = - \dfrac{1}{4}\\ \dfrac{x}{2} = \dfrac{{ - 5}}{4} \end{array} \right.\\ ⇔ \left[ \begin{array}{l} x = \dfrac{{ - 1}}{2}\\ x = - \dfrac{5}{2} \end{array} \right.\\ d){\left( {\dfrac{5}{6} - \dfrac{{3x}}{2}} \right)^5} = \dfrac{{243}}{{1024}}\\ ⇔ \dfrac{5}{6} - \dfrac{{3x}}{2} = \dfrac{3}{4}\\ ⇔ \dfrac{{3x}}{2} = \dfrac{5}{6} - \dfrac{3}{4}\\ ⇔ \dfrac{{3x}}{2} = \dfrac{1}{{12}}\\ ⇔ x = \dfrac{1}{{18}}$