\(\begin{array}{l}5 - \left| {\frac{3}{2} + x} \right| = \frac{1}{2}\\ \Leftrightarrow \,\left| {\frac{3}{2} + x} \right| = 5 - \frac{1}{2}\\ \Leftrightarrow \left| {\frac{3}{2} + x} \right| = \frac{9}{2}\end{array}\)
\(\begin{array}{l}TH1:\,\,\frac{3}{2} + x \ge 0 \Leftrightarrow x \ge \frac{{ - 3}}{2}\\ \Rightarrow \left| {\frac{3}{2} + x} \right| = \frac{9}{2}\\ \Leftrightarrow \frac{3}{2} + x = \frac{9}{2}\\ \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,x = \frac{9}{2} - \frac{3}{2}\\ \Leftrightarrow \,\,\,\,\,\,\,\,\,\,x = \,\,\,3\,\,\,\left( {TM} \right)\\TH2:\,\,\frac{3}{2} + x < 0 \Leftrightarrow x < \frac{{ - 3}}{2}\\ \Rightarrow \left| {\frac{3}{2} + x} \right| = \frac{9}{2}\\ \Leftrightarrow \frac{3}{2} + x = \frac{{ - 9}}{2}\\ \Leftrightarrow \,\,\,\,\,\,\,\,\,x = \frac{{ - 9}}{2} - \frac{3}{2}\\ \Leftrightarrow \,\,\,\,\,\,\,\,\,x = - 6\,\,\,\,\left( {TM} \right)\end{array}\)
