Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
D = \left[ {5;9} \right]\\
\sqrt {x - 5}  - \sqrt {9 - x}  > 1\\
 \Leftrightarrow \sqrt {x - 5}  > \sqrt {9 - x}  + 1\\
 \Leftrightarrow x - 5 > {\left( {\sqrt {9 - x}  + 1} \right)^2}\\
 \Leftrightarrow x - 5 > 9 - x + 2\sqrt {9 - x}  + 1\\
 \Leftrightarrow 2x - 15 > 2\sqrt {9 - x} \\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - 15 > 0\\
{\left( {2x - 15} \right)^2} > 4\left( {9 - x} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x > \dfrac{{15}}{2}\\
4{x^2} - 56x + 189 > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x > \dfrac{{15}}{2}\\
\left[ \begin{array}{l}
x > \dfrac{{14 + \sqrt 7 }}{2}\\
x < \dfrac{{14 - \sqrt 7 }}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow x > \dfrac{{14 + \sqrt 7 }}{2}\\
 \Rightarrow S = \left( {\dfrac{{14 + \sqrt 7 }}{2};9} \right]\\
b,\\
\sqrt {{x^2} - x - 12}  < 7 - x\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - x - 12 \ge 0\\
7 - x > 0\\
{x^2} - x - 12 < {\left( {7 - x} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 4} \right)\left( {x + 3} \right) \ge 0\\
x < 7\\
{x^2} - x - 12 < 49 - 14x + {x^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 4\\
x \le  - 3
\end{array} \right.\\
x < 7\\
x < \dfrac{{61}}{{13}}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4 \le x < \dfrac{{61}}{{13}}\\
x \le  - 3
\end{array} \right.\\
 \Rightarrow S = \left( { - \infty ; - 3} \right] \cup \left[ {4;\dfrac{{61}}{{13}}} \right)\\
c,\\
\sqrt { - {x^2} - 4x + 21}  < x + 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - {x^2} - 4x + 21 \ge 0\\
x + 3 > 0\\
 - {x^2} - 4x + 21 < {x^2} + 6x + 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 4x - 21 \le 0\\
x >  - 3\\
2{x^2} + 10x - 12 > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 7} \right)\left( {x - 3} \right) \le 0\\
x >  - 3\\
2\left( {x - 1} \right)\left( {x + 6} \right) > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 7 \le x \le 3\\
x >  - 3\\
\left[ \begin{array}{l}
x > 1\\
x <  - 6
\end{array} \right.
\end{array} \right. \Leftrightarrow 1 < x \le 3\\
 \Rightarrow S = \left( {1;3} \right]\\
d,\\
\sqrt {{x^2} - 3x - 10}  \ge x - 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x - 10 \ge 0\\
\left[ \begin{array}{l}
x - 2 \le 0\\
\left\{ \begin{array}{l}
x - 2 > 0\\
{x^2} - 3x - 10 \ge {x^2} - 4x + 4
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 5} \right)\left( {x + 2} \right) \ge 0\\
\left[ \begin{array}{l}
x \le 2\\
\left\{ \begin{array}{l}
x > 2\\
x \ge 14
\end{array} \right.
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 5\\
x \le  - 2
\end{array} \right.\\
\left[ \begin{array}{l}
x \le 2\\
x \ge 14
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 14\\
x \le  - 2
\end{array} \right.\\
 \Rightarrow S = \left( { - \infty ; - 2} \right] \cup \left[ {14; + \infty } \right)
\end{array}\)