Đáp án:

$\begin{array}{l}
e)2x > \sqrt {{x^2} - 3x - 10} \\
 \Rightarrow \left\{ \begin{array}{l}
x > 0\\
{x^2} - 3x - 10 \ge 0\\
4{x^2} > {x^2} - 3x - 10
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x > 0\\
\left( {x - 5} \right)\left( {x + 2} \right) \ge 0\\
3{x^2} + 3x + 10 > 0\left( {tm} \right)
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x > 0\\
\left[ \begin{array}{l}
x \ge 5\\
x \le  - 2
\end{array} \right.
\end{array} \right.\\
 \Rightarrow x \ge 5\\
Vay\,x \ge 5\\
g)\left( {x - 1} \right)\sqrt {{x^2} - x - 1}  \le 0\\
 \Rightarrow \left\{ \begin{array}{l}
x - 1 \le 0\\
{x^2} - x - 1 \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x \le 1\\
{\left( {x - \frac{1}{2}} \right)^2} \ge \frac{5}{4}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x \ge \frac{{\sqrt 5  + 1}}{2}\\
x \le \frac{{ - \sqrt 5  + 1}}{2}
\end{array} \right.
\end{array} \right.\\
 \Rightarrow x \le \frac{{1 - \sqrt 5 }}{2}\\
f)\left( {{x^2} - 3x} \right)\sqrt {2{x^2} - 3x - 2}  \ge 0\\
 \Rightarrow x\left( {x - 3} \right)\sqrt {2{x^2} - 3x - 2}  \ge 0\\
 \Rightarrow \left\{ \begin{array}{l}
x\left( {x - 3} \right) \ge 0\\
2{x^2} - 3x - 2 \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le 0
\end{array} \right.\\
\left( {2x + 1} \right)\left( {x - 2} \right) \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le 0
\end{array} \right.\\
\left[ \begin{array}{l}
x \ge 2\\
x \le  - \frac{1}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le  - \frac{1}{2}
\end{array} \right.\\
Vay\,x \ge 3;x \le  - \frac{1}{2}\\
k)\left( {x - 2} \right)\sqrt {{x^2} + 4}  < {x^2} - 4\\
 \Rightarrow \left( {x - 2} \right)\sqrt {{x^2} + 4}  - \left( {x - 2} \right)\left( {x + 2} \right) < 0\\
 \Rightarrow \left( {x - 2} \right)\left( {\sqrt {{x^2} + 4}  - x - 2} \right) < 0\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 2\\
\sqrt {{x^2} + 4}  > x + 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x > 2\\
\sqrt {{x^2} + 4}  < x + 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 2\\
{x^2} + 4 > {x^2} + 4x + 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x > 2\\
{x^2} + 4 < {x^2} + 4x + 4
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x < 0\\
x > 2
\end{array} \right.\\
l)3{x^2} - \left| {5x + 2} \right| > 0\\
 \Rightarrow 3{x^2} > \left| {5x + 2} \right|\\
 \Rightarrow \left[ \begin{array}{l}
3{x^2} > 5x + 2\\
3{x^2} <  - 5x - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\left( {3x + 1} \right)\left( {x - 2} \right) > 0\\
\left( {3x + 2} \right)\left( {x + 1} \right) < 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x > 2\\
x <  - \frac{1}{3}\\
 - 1 < x <  - \frac{2}{3}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x > 2\\
 - 1 < x <  - \frac{2}{3}
\end{array} \right.
\end{array}$