Đáp án:
$\begin{array}{l}
\frac{1}{{x - 2}} - \frac{1}{x} - \frac{2}{{x + 2}} \le 0\\
\Rightarrow \frac{{x\left( {x + 2} \right) - \left( {x - 2} \right)\left( {x + 2} \right) - 2x\left( {x - 2} \right)}}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} \le 0\\
\Rightarrow \frac{{{x^2} + 2x - {x^2} + 4 - 2{x^2} + 4x}}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} \le 0\\
\Rightarrow \frac{{3x + 2}}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + 2 \le 0\\
x\left( {{x^2} - 4} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 2 \ge 0\\
x\left( {{x^2} - 4} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
- 2 < x \le - \frac{2}{3}\\
0 < x < 2
\end{array} \right.
\end{array}$
