Đáp án:
$\begin{array}{l}
a)\left| x \right| = 3,2\\
\Leftrightarrow x = 3,2;x = - 3,2\\
Vậy\,x = 3,2;x = - 3,2\\
b)\left| x \right| = \dfrac{2}{7}\\
\Leftrightarrow x = - \dfrac{2}{7}\left( {x < 0} \right)\\
Vậy\,x = - \dfrac{2}{7}\\
c)\left| x \right| = - 2\dfrac{5}{9}\left( {ktm:do:\left| x \right| > 0} \right)\\
Vậy\,x \in \emptyset \\
d)\left| x \right| = 0,54\\
\Leftrightarrow x = 0,54\left( {do:x > 0} \right)\\
Vậy\,x = 0,54\\
e)\left| x \right| = - \dfrac{8}{{21}}\left( {ktm:do:\left| x \right| > 0} \right)\\
Vậy\,x \in \emptyset \\
f)\left| {5\dfrac{1}{2} - 2x} \right| = 1\dfrac{6}{7}\\
\Leftrightarrow \left| {\dfrac{{11}}{2} - 2x} \right| = \dfrac{{13}}{7}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{11}}{2} - 2x = \dfrac{{13}}{7}\\
\dfrac{{11}}{2} - 2x = - \dfrac{{13}}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{11}}{2} - \dfrac{{13}}{7} = \dfrac{{51}}{{14}}\\
2x = \dfrac{{11}}{2} + \dfrac{{13}}{7} = \dfrac{{103}}{{14}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{51}}{{28}}\\
x = \dfrac{{103}}{{28}}
\end{array} \right.\\
Vậy\,x = \dfrac{{51}}{{28}};x = \dfrac{{103}}{{28}}\\
g)\left| {x - 0,25} \right| = \dfrac{{15}}{6}\\
\Leftrightarrow \left| {x - 0,25} \right| = 2,5\\
\Leftrightarrow \left[ \begin{array}{l}
x - 0,25 = 2,5\\
x - 0,25 = - 2,5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2,75\\
x = - 2,25
\end{array} \right.\\
Vậy\,x = 2,75;x = - 2,25\\
h) - 6\left| {x + 0,4} \right| = - 24\\
\Leftrightarrow \left| {x + 0,4} \right| = 4\\
\Leftrightarrow \left[ \begin{array}{l}
x + 0,4 = 4\\
x + 0,4 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3,6\\
x = - 4,4
\end{array} \right.\\
Vậy\,x = 3,6;x = - 4,4
\end{array}$
