Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
\left| {2x - \dfrac{1}{3}} \right| = \dfrac{2}{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = \dfrac{2}{3}\\
2x - \dfrac{1}{3} =  - \dfrac{2}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
2x =  - \dfrac{1}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{1}{6}
\end{array} \right.\\
b,\\
\left| {\dfrac{1}{2}x + \dfrac{1}{3}} \right| = \dfrac{3}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x + \dfrac{1}{3} = \dfrac{3}{2}\\
\dfrac{1}{2}x + \dfrac{1}{3} =  - \dfrac{3}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x = \dfrac{7}{6}\\
\dfrac{1}{2}x =  - \dfrac{{11}}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x =  - \dfrac{{11}}{3}
\end{array} \right.\\
c,\\
\dfrac{1}{2} - \left| {\dfrac{1}{3} - x} \right| = \dfrac{1}{4}\\
 \Leftrightarrow \left| {\dfrac{1}{3} - x} \right| = \dfrac{1}{2} - \dfrac{1}{4}\\
 \Leftrightarrow \left| {\dfrac{1}{3} - x} \right| = \dfrac{1}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{3} - x = \dfrac{1}{4}\\
\dfrac{1}{3} - x =  - \dfrac{1}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{{12}}\\
x = \dfrac{7}{{12}}
\end{array} \right.\\
d,\\
\dfrac{1}{2}.\left| {2x - \dfrac{1}{4}} \right| = \dfrac{1}{3}\\
 \Leftrightarrow \left| {2x - \dfrac{1}{4}} \right| = \dfrac{2}{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{4} = \dfrac{2}{3}\\
2x - \dfrac{1}{4} =  - \dfrac{2}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{11}}{{12}}\\
2x =  - \dfrac{5}{{12}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{{24}}\\
x =  - \dfrac{5}{{24}}
\end{array} \right.\\
2,\\
a,\\
\left| {x + \dfrac{1}{4}} \right| = 0\\
 \Leftrightarrow x + \dfrac{1}{4} = 0\\
 \Leftrightarrow x =  - \dfrac{1}{4}\\
b,\\
\left| {x - \dfrac{2}{5}} \right| \ge 0,\,\,\,\forall x \Rightarrow \left| {x - \dfrac{2}{5}} \right| >  - \dfrac{1}{2}
\end{array}\)