$\begin{array}{l} \left( {{x^2} - 4x - 5} \right)\sqrt {4 - {x^2}} \ge 0\left( { - 2 \le x \le 2} \right)\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 4x - 5 \ge 0\\ \sqrt {4 - {x^2}} \ge 0 \end{array} \right. hoặc \left\{ \begin{array}{l} {x^2} - 4x - 5 \le 0\\ \sqrt {4 - {x^2}} \le 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x \ge 5\\ x \le - 1 \end{array} \right.\\ - 2 \le x \le 2 \end{array} \right. hoặc \left\{ \begin{array}{l} - 1 \le x \le 5\\ \left[ \begin{array}{l} x = 2\\ x = - 2 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 2 \le x \le 1\\ x = 2 \end{array} \right.\\ S = \left[ { - 2;1} \right] \cup \left\{ 2 \right\} \end{array}$
