Đáp án:
$\begin{array}{l}
\dfrac{{{x^2} - 36}}{{x - 5}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 36 \le 0\\
x - 5 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 36 \ge 0\\
x - 5 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} \le 36\\
x > 5
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} \ge 36\\
x < 5
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 6 \le x \le 6\\
x > 5
\end{array} \right.\\
x \le - 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
5 < x \le 6\\
x \le - 6
\end{array} \right.\\
Vậy\,5 < x \le 6;x \le - 6
\end{array}$
