Đáp án:
$\begin{array}{l}
a)\left| {3x - 2} \right| > 7\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 2 > 7\\
3x - 2 < - 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x > 9\\
3x < - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x < - \dfrac{5}{3}
\end{array} \right.\\
Vậy\,x > 3\,hoặc\,x < - \dfrac{5}{3}\\
b)\left| {5x - 12} \right| < 3\\
\Leftrightarrow - 3 < 5x - 12 < 3\\
\Leftrightarrow 9 < 5x < 15\\
\Leftrightarrow \dfrac{9}{5} < x < 3\\
Vậy\,\dfrac{9}{5} < x < 3\\
c)\left| {2x - 8} \right| \le 7\\
\Leftrightarrow - 7 \le 2x - 8 \le 7\\
\Leftrightarrow 1 \le 2x \le 15\\
\Leftrightarrow \dfrac{1}{2} \le x \le \dfrac{{15}}{2}\\
Vậy\,\dfrac{1}{2} \le x \le \dfrac{{15}}{2}\\
d)\left| {3x + 15} \right| \ge 3\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 15 \ge 3\\
3x + 15 \le - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge - 4\\
x \le - 6
\end{array} \right.\\
e)\left| {x - 1} \right| > \dfrac{{x + 1}}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 > \dfrac{{x + 1}}{2}\\
x - 1 < \dfrac{{ - x - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 2 > x + 1\\
2x - 2 < - x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x < \dfrac{1}{3}
\end{array} \right.\\
f)\left| {x - 2} \right| < \dfrac{x}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 < \dfrac{x}{2}\\
x - 2 > - \dfrac{x}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > - 4\\
x > \dfrac{4}{3}
\end{array} \right.\\
\Leftrightarrow x > \dfrac{4}{3}\\
h)\left| {2x + 1} \right| \le x\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 \le x\\
2x + 1 \ge - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 1\\
x \ge - \dfrac{1}{3}
\end{array} \right.\\
i)\left| {x - 2} \right| > x + 1\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 > x + 1\\
x - 2 < - x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 2 > 1\left( {ktm} \right)\\
x < \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow x < \dfrac{1}{2}
\end{array}$
