Đáp án:
$\begin{array}{l}
c)\left| {x - 1,5} \right| + \left| {2,5 - x} \right| = 0\\
Do:\left\{ \begin{array}{l}
\left| {x - 1,5} \right| \ge 0\\
\left| {2,5 - x} \right| \ge 0
\end{array} \right.\\
\Leftrightarrow \left| {x - 1,5} \right| + \left| {2,5 - x} \right| \ge 0\\
Khi:\left| {x - 1,5} \right| + \left| {2,5 - x} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1,5 = 0\\
2,5 - x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1,5\\
x = 2,5
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
d)\left| {x - \dfrac{4}{5}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{4}{5} = \dfrac{3}{4}\\
x - \dfrac{4}{5} = - \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{4} + \dfrac{4}{5} = \dfrac{{31}}{{20}}\\
x = \dfrac{4}{5} - \dfrac{3}{4} = \dfrac{1}{{20}}
\end{array} \right.\\
Vậy\,x = \dfrac{{31}}{{20}};x = \dfrac{1}{{20}}
\end{array}$
