Đáp án:
$\begin{array}{l}
\left| {\frac{{2x - 7}}{{x + 1}}} \right| \ge 3\\
\Rightarrow \left[ \begin{array}{l}
\frac{{2x - 7}}{{x + 1}} \ge 3\\
\frac{{2x - 7}}{{x + 1}} \le - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\frac{{2x - 7 - 3\left( {x + 1} \right)}}{{x + 1}} \ge 0\\
\frac{{2x - 7 + 3\left( {x + 1} \right)}}{{x + 1}} \le 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\frac{{ - x - 10}}{{x + 1}} \ge 0\\
\frac{{5x - 4}}{{x + 1}} \le 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
- 10 \le x < - 1\\
- 1 < x \le \frac{4}{5}
\end{array} \right.\\
Vậy\, - 10 \le x \le \frac{4}{5};x \ne - 1
\end{array}$
