ét o ét cứu mình vs mình đâng cần gấp huhu tks="(

Đáp án:

$\begin{array}{l}
a)\left| { - 2x - 1,5} \right| = \dfrac{1}{2}\\
 \Leftrightarrow \left| {2x + 1,5} \right| = 0,5\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1,5 = 0,5\\
2x + 1,5 =  - 0,5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 0,5 - 1,5\\
2x =  - 0,5 - 1,5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - 1\\
2x =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x =  - 1
\end{array} \right.\\
Vậy\,x =  - \dfrac{1}{2};x =  - 1\\
b) - \dfrac{3}{2} - \left| {2,5 - x} \right| =  - \dfrac{5}{3}\\
 \Leftrightarrow \left| {\dfrac{5}{2} - x} \right| =  - \dfrac{3}{2} + \dfrac{5}{3}\\
 \Leftrightarrow \left| {\dfrac{5}{2} - x} \right| = \dfrac{1}{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{5}{2} - x = \dfrac{1}{6}\\
\dfrac{5}{2} - x =  - \dfrac{1}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2} - \dfrac{1}{6} = \dfrac{7}{3}\\
x = \dfrac{5}{2} + \dfrac{1}{6} = \dfrac{8}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{3};x = \dfrac{8}{3}\\
c)\left| {2x - 1} \right| = \left| {3x - 2,5} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 3x - 2,5\\
2x - 1 = 2,5 - 3x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2,5 - 1 = 1,5\\
2x + 3x = 2,5 + 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1,5\\
5x = 3,5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1,5\\
x = 0,7
\end{array} \right.\\
Vậy\,x = 1,5;x = 0,7\\
d)\left| {\dfrac{2}{5}x - 3} \right| + \dfrac{3}{5}x = \dfrac{2}{3}\\
 \Leftrightarrow \left| {\dfrac{2}{5}x - 3} \right| = \dfrac{2}{3} - \dfrac{3}{5}x\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5}x - 3 = \dfrac{2}{3} - \dfrac{3}{5}x\\
\dfrac{2}{5}x - 3 = \dfrac{3}{5}x - \dfrac{2}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5}x + \dfrac{3}{5}x = \dfrac{2}{3} + 3\\
\dfrac{2}{5}x - \dfrac{3}{5}x = 3 - \dfrac{2}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{3}\\
 - \dfrac{1}{5}x = \dfrac{7}{3} \Leftrightarrow x =  - \dfrac{{35}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{11}}{3};x =  - \dfrac{{35}}{3}
\end{array}$