Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left| {3x - 5} \right| < 2\\
\Leftrightarrow - 2 < 3x - 5 < 2\\
\Leftrightarrow - 2 + 5 < 3x < 2 + 5\\
\Leftrightarrow 3 < 3x < 7\\
\Leftrightarrow 1 < x < \dfrac{7}{3}\\
\Rightarrow S = \left( {1;\dfrac{7}{3}} \right)\\
b,\\
\left| {\dfrac{{2 - x}}{{x + 1}}} \right| \ge 2\,\,\,\,\,\,\,\,\,\,\left( {x \ne - 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{2 - x}}{{x + 1}} \ge 2\\
\dfrac{{2 - x}}{{x + 1}} \le - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{2 - x}}{{x + 1}} - 2 \ge 0\\
\dfrac{{2 - x}}{{x + 1}} + 2 \le 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{2 - x - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)}} \ge 0\\
\dfrac{{2 - x + 2\left( {x + 1} \right)}}{{x + 1}} \le 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ - 3x}}{{x + 1}} \ge 0\\
\dfrac{{x + 4}}{{x + 1}} \le 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{{x + 1}} \le 0\\
\dfrac{{x + 4}}{{x + 1}} \le 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
- 1 < x \le 0\\
- 4 \le x < - 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
- 4 \le x \le 0\\
x \ne - 1
\end{array} \right.
\end{array}\)
