Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\left| {x + \dfrac{3}{5}} \right| - \left| {x - \dfrac{7}{3}} \right| = 0\\
 \Leftrightarrow \left| {x + \dfrac{3}{5}} \right| = \left| {x - \dfrac{7}{3}} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{5} = x - \dfrac{7}{3}\\
x + \dfrac{3}{5} = \dfrac{7}{3} - x
\end{array} \right. \Leftrightarrow x = \dfrac{{13}}{{15}}\\
2,\\
\left| {x - 2} \right| < 3\\
 \Leftrightarrow  - 3 < x - 2 < 3\\
 \Leftrightarrow  - 1 < x < 5\\
3,\\
\left| {\dfrac{5}{3}x} \right| = \left| { - \dfrac{1}{6}} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{5}{3}x =  - \dfrac{1}{6}\\
\dfrac{5}{3}x = \dfrac{1}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{{10}}\\
x = \dfrac{1}{{10}}
\end{array} \right.\\
4,\\
\left| {\dfrac{1}{2}x} \right| = 3x - 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
3x - 2 \ge 0\\
\left[ \begin{array}{l}
\dfrac{1}{2}x = 3x - 2\\
\dfrac{1}{2}x =  - 3x + 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{2}{3}\\
\left[ \begin{array}{l}
x = \dfrac{4}{5}\\
x = \dfrac{4}{7}
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{4}{5}\\
5,\\
\left| {x - 1} \right| = 3x + 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
3x + 2 \ge 0\\
\left[ \begin{array}{l}
x - 1 = 3x + 2\\
x - 1 =  - 3x - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{2}{3}\\
\left[ \begin{array}{l}
x =  - \dfrac{3}{2}\\
x =  - \dfrac{1}{4}
\end{array} \right.
\end{array} \right. \Leftrightarrow x =  - \dfrac{1}{4}\\
6,\\
\left| {5x} \right| = x - 12\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 12 \ge 0\\
\left[ \begin{array}{l}
5x = x - 12\\
5x =  - x + 12
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 12\\
\left[ \begin{array}{l}
x =  - 3\\
x = 2
\end{array} \right.
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
7,\\
\left| {7 - x} \right| = 5x + 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
5x + 1 \ge 0\\
\left[ \begin{array}{l}
7 - x = 5x + 1\\
7 - x =  - 5x - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{5}\\
\left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 1\\
8,\\
\left| {\dfrac{3}{4}x - \dfrac{3}{4}} \right| - \dfrac{3}{4} = \left| { - \dfrac{3}{4}} \right|\\
 \Leftrightarrow \left| {\dfrac{3}{4}x - \dfrac{3}{4}} \right| - \dfrac{3}{4} = \dfrac{3}{4}\\
 \Leftrightarrow \left| {\dfrac{3}{4}x - \dfrac{3}{4}} \right| = \dfrac{3}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{4}x - \dfrac{3}{4} = \dfrac{3}{2}\\
\dfrac{3}{4}x - \dfrac{3}{4} =  - \dfrac{3}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 1
\end{array} \right.
\end{array}\)