Câu 1:
$\begin{array}{l} \left| {2{x^2} - 3x - 15} \right| \le - 2{x^2} - 8x - 6\\ \Leftrightarrow 2{x^2} + 8x + 6 \le 2{x^2} - 3x - 15 \le - 2{x^2} - 8x - 6\\ \Leftrightarrow \left\{ \begin{array}{l} 11x \le - 21\\ 4{x^2} + 5x - 9 \le 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \le \dfrac{{ - 21}}{{11}}\\ \dfrac{{ - 9}}{4} \le x \le 1 \end{array} \right. \Rightarrow \dfrac{{ - 9}}{4} \le x \le \dfrac{{ - 21}}{{11}}\\ \Rightarrow S = \left[ {\dfrac{{ - 9}}{4};\dfrac{{ - 21}}{{11}}} \right] \end{array}$
Câu 2:
$\begin{array}{l}
\dfrac{{3\left( {x + 1} \right) - \sqrt {2{x^2} - 3x + 1} }}{{2\left( {{x^2} + 1} \right)}} \ge 1\\
\Leftrightarrow \dfrac{{3\left( {x + 1} \right) - \sqrt {2{x^2} - 3x + 1} - 2{x^2} - 2}}{{2\left( {{x^2} + 1} \right)}} \ge 0\\
\Leftrightarrow \dfrac{{ - 2{x^2} + 3x + 1 - \sqrt {2{x^2} - 3x + 1} }}{{2\left( {{x^2} + 1} \right)}} \ge 0\\
\Leftrightarrow - 2{x^2} + 3x + 1 - \sqrt {2{x^2} - 3x + 1} \ge 0\\
\Leftrightarrow 2{x^2} - 3x - 1 + \sqrt {2{x^2} - 3x + 1} \le 0\\
\Leftrightarrow {t^2} + t - 2 \le 0\left( {t = \sqrt {2{x^2} - 3x + 1} \ge 0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
t \ge 0\\
- 2 \le x \le 1
\end{array} \right. \Rightarrow 0 \le t \le 1\\
\Rightarrow 0 \le \sqrt {2{x^2} - 3x + 1} \le 1\\
\Leftrightarrow 0 \le 2{x^2} - 3x + 1 \le 1\\
\Leftrightarrow \left\{ \begin{array}{l}
0 \le x \le \dfrac{3}{2}\\
\left[ \begin{array}{l}
x \le \dfrac{1}{2}\\
x \ge 1
\end{array} \right.
\end{array} \right. \Rightarrow S = \left[ { 0 ;\dfrac{1}{2}} \right] \cup \left[ {1;\dfrac{3}{2}} \right]\\
\end{array}$
