Đáp án:$\left[ \begin{array}{l}
- 1 \le x < 2\\
x < - 2
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\frac{1}{{x - 2}} - \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \frac{{x + 2 - 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \frac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
\left( {x - 2} \right)\left( {x + 2} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
\left( {x - 2} \right)\left( {x + 2} \right) > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 1\\
- 2 < x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - 1\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
- 1 \le x < 2\\
x < - 2
\end{array} \right.
\end{array}$
