Đáp án:

\(\begin{array}{l}
C1:\\
a)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x =  - 2\\
x =  - 6
\end{array} \right.\\
b)x \in \emptyset \\
d)\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
C2:\\
a)\left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\\
c)x = \dfrac{1}{6}\\
b)\left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x =  - \dfrac{5}{6}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = \dfrac{{65}}{8}\\
x =  - \dfrac{{55}}{{16}}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
C1:\\
a)\left| x \right| = \dfrac{3}{2}\\
 \to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
c)\left| {x + 4} \right| = 2\\
 \to \left[ \begin{array}{l}
x + 4 = 2\\
x + 4 =  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2\\
x =  - 6
\end{array} \right.\\
b)\left| {\dfrac{3}{2}x} \right| =  - 2\left( {KTM} \right)\\
Do:\left| {\dfrac{3}{2}x} \right| \ge 0\forall x\\
 \to x \in \emptyset \\
d)\left| {2x - 4} \right| = 4\\
 \to \left[ \begin{array}{l}
2x - 4 = 4\\
2x - 4 =  - 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
C2:\\
a)\left| {2x - 3} \right| = \dfrac{1}{3}\\
 \to \left[ \begin{array}{l}
2x - 3 = \dfrac{1}{3}\left( {DK:x \ge \dfrac{3}{2}} \right)\\
2x - 3 =  - \dfrac{1}{3}\left( {DK:x < \dfrac{3}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{{10}}{3}\\
2x = \dfrac{8}{3}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\left( {TM} \right)\\
c)\left| {x - 1} \right| - 2x = \dfrac{1}{2}\\
 \to \left| {x - 1} \right| = 2x + \dfrac{1}{2}\\
 \to \left[ \begin{array}{l}
x - 1 = 2x + \dfrac{1}{2}\left( {DK:x \ge 1} \right)\\
x - 1 =  - 2x - \dfrac{1}{2}\left( {DK:x < 1} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{3}{2}\left( l \right)\\
3x = \dfrac{1}{2}
\end{array} \right.\\
 \to x = \dfrac{1}{6}\\
b)\left| {x + \dfrac{1}{4}} \right| = \dfrac{5}{6} - \dfrac{1}{4}\\
 \to \left| {x + \dfrac{1}{4}} \right| = \dfrac{7}{{12}}\\
 \to \left[ \begin{array}{l}
x + \dfrac{1}{4} = \dfrac{7}{{12}}\\
x + \dfrac{1}{4} =  - \dfrac{7}{{12}}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x =  - \dfrac{5}{6}
\end{array} \right.\\
d)3x - \dfrac{5}{4} = \left| {x + 15} \right|\\
 \to \left[ \begin{array}{l}
3x - \dfrac{5}{4} = x + 15\\
3x - \dfrac{5}{4} =  - x - 15
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{{65}}{4}\\
4x =  - \dfrac{{55}}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{65}}{8}\\
x =  - \dfrac{{55}}{{16}}
\end{array} \right.
\end{array}\)