Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = \dfrac{5}{8}\\
x =  - \dfrac{1}{8}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{{11}}{{180}}\\
x =  - \dfrac{{11}}{{180}}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{1}{8}\\
x =  - \dfrac{9}{8}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = 0\\
x = \dfrac{3}{4}\\
x =  - \dfrac{3}{4}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\left| {2x - \dfrac{1}{2}} \right| = \dfrac{3}{4}\\
 \to \left[ \begin{array}{l}
2x - \dfrac{1}{2} = \dfrac{3}{4}\left( {DK:x \ge \dfrac{1}{4}} \right)\\
2x - \dfrac{1}{2} =  - \dfrac{3}{4}\left( {DK:x < \dfrac{1}{4}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{5}{4}\\
2x =  - \dfrac{1}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{8}\\
x =  - \dfrac{1}{8}
\end{array} \right.\\
b)\left| x \right| - \dfrac{{38}}{7}\left| x \right| - \dfrac{3}{4} = 2\left| x \right| - \dfrac{8}{7}\\
 \to \dfrac{{45}}{7}\left| x \right| = \dfrac{{11}}{{28}}\\
 \to \left| x \right| = \dfrac{{11}}{{180}}\\
 \to \left[ \begin{array}{l}
x = \dfrac{{11}}{{180}}\\
x =  - \dfrac{{11}}{{180}}
\end{array} \right.\\
b)\dfrac{4}{{13}}\left( {\dfrac{{6 + 2}}{5}} \right).{\left( {2x + 1} \right)^2} = \dfrac{{10}}{{13}}\\
 \to \dfrac{{32}}{{65}}.{\left( {2x + 1} \right)^2} = \dfrac{{10}}{{13}}\\
 \to {\left( {2x + 1} \right)^2} = \dfrac{{25}}{{16}}\\
 \to \left| {2x + 1} \right| = \dfrac{5}{4}\\
 \to \left[ \begin{array}{l}
2x + 1 = \dfrac{5}{4}\left( {DK:x \ge  - \dfrac{1}{2}} \right)\\
2x + 1 =  - \dfrac{5}{4}\left( {DK:x <  - \dfrac{1}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{1}{4}\\
2x =  - \dfrac{9}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{8}\\
x =  - \dfrac{9}{8}
\end{array} \right.\\
d)x\left( {{x^2} - \dfrac{9}{{16}}} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
{x^2} - \dfrac{9}{{16}} = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{3}{4}\\
x =  - \dfrac{3}{4}
\end{array} \right.
\end{array}\)