Đáp án:

$\begin{array}{l}
1)\dfrac{3}{4} - \left| {2x + 1} \right| = \dfrac{7}{8}\\
 \Leftrightarrow \left| {2x + 1} \right| = \dfrac{3}{4} - \dfrac{7}{8}\\
 \Leftrightarrow \left| {2x + 1} \right| = \dfrac{{ - 1}}{8}\left( {ktm:do:\left| {2x + 1} \right| \ge 0} \right)\\
Vậy\,x \in \emptyset \\
2)\dfrac{{21}}{5} + 3:\left| {\dfrac{x}{4} - \dfrac{2}{3}} \right| = 6\\
 \Leftrightarrow 3:\left| {\dfrac{x}{4} - \dfrac{2}{3}} \right| = 6 - \dfrac{{21}}{5}\\
 \Leftrightarrow 3:\left| {\dfrac{x}{4} - \dfrac{2}{3}} \right| = \dfrac{9}{5}\\
 \Leftrightarrow \left| {\dfrac{x}{4} - \dfrac{2}{3}} \right| = 3:\dfrac{9}{5}\\
 \Leftrightarrow \left| {\dfrac{x}{4} - \dfrac{2}{3}} \right| = \dfrac{5}{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{4} - \dfrac{2}{3} = \dfrac{5}{3} \Leftrightarrow \dfrac{x}{4} = \dfrac{7}{3} \Leftrightarrow x = \dfrac{{28}}{3}\\
\dfrac{x}{4} - \dfrac{2}{3} =  - \dfrac{5}{3} \Leftrightarrow \dfrac{x}{4} =  - 1 \Leftrightarrow x =  - 4
\end{array} \right.\\
Vậy\,x = \dfrac{{28}}{3};x =  - 4\\
3)\left| {\dfrac{7}{8}x + \dfrac{5}{6}} \right| - \left| {\dfrac{1}{2}x + 5} \right| = 0\\
 \Leftrightarrow \left| {\dfrac{7}{8}x + \dfrac{5}{6}} \right| = \left| {\dfrac{1}{2}x + 5} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{7}{8}x + \dfrac{5}{6} = \dfrac{1}{2}x + 5\\
\dfrac{7}{8}x + \dfrac{5}{6} =  - \dfrac{1}{2}x - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{7}{8}x - \dfrac{1}{2}x = 5 - \dfrac{5}{6}\\
\dfrac{7}{8}x + \dfrac{1}{2}x =  - 5 - \dfrac{5}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{8}x = \dfrac{{25}}{6} \Leftrightarrow x = \dfrac{{25}}{6}.\dfrac{8}{3} = \dfrac{{100}}{9}\\
\dfrac{{11}}{8}x = \dfrac{{ - 35}}{6} \Leftrightarrow x = \dfrac{{ - 35}}{6}.\dfrac{8}{{11}} = \dfrac{{ - 140}}{{33}}
\end{array} \right.\\
Vậy\,x = \dfrac{{100}}{9};x = \dfrac{{ - 140}}{{33}}\\
4)\left| {x - 1} \right| = 3x + 2\left( {dk:x \ge  - \dfrac{2}{3}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 3x + 2\\
x - 1 =  - 3x - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - 3 \Leftrightarrow x = \dfrac{{ - 3}}{2}\left( {ktm} \right)\\
4x =  - 1 \Leftrightarrow x = \dfrac{{ - 1}}{4}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - \dfrac{1}{4}\\
5)\left| {2{x^2} + 1} \right| = 4x + 1\left( {dk:x \ge  - \dfrac{1}{4}} \right)\\
 \Leftrightarrow 2{x^2} + 1 = 4x + 1\\
 \Leftrightarrow 2{x^2} - 4x = 0\\
 \Leftrightarrow 2x\left( {x - 2} \right) = 0\\
 \Leftrightarrow x = 2;x = 0\left( {tmdk} \right)\\
Vậy\,x = 2;x = 0
\end{array}$