Đáp án:
$\begin{array}{l}
8)Dkxd:\dfrac{{x - 1}}{{5x}} > 0 \Rightarrow \left[ \begin{array}{l}
x > 1\\
x < 0
\end{array} \right.\\
Dat:\sqrt {\dfrac{{x - 1}}{{5x}}} = t\left( {t > 0} \right)\\
\Rightarrow \sqrt {\dfrac{{5x}}{{x - 1}}} = \dfrac{1}{t}\\
\Rightarrow 4 < 21.t - \dfrac{1}{t}\\
\Rightarrow 21{t^2} - 4t - 1 > 0\left( {do:t > 0} \right)\\
\Rightarrow 21{t^2} - 7t + 3t - 1 > 0\\
\Rightarrow \left( {3t - 1} \right)\left( {7t + 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
t > \dfrac{1}{3}\\
t < - \dfrac{1}{7}
\end{array} \right.\\
\Rightarrow t > \dfrac{1}{3}\left( {do:t > 0} \right)\\
\Rightarrow \sqrt {\dfrac{{x - 1}}{{5x}}} > \dfrac{1}{3}\\
\Rightarrow \dfrac{{x - 1}}{{5x}} > \dfrac{1}{9}\\
\Rightarrow \dfrac{{9\left( {x - 1} \right) - 5x}}{{45x}} > 0\\
\Rightarrow \dfrac{{4x - 9}}{{45x}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > \dfrac{9}{4}\\
x < 0
\end{array} \right.\\
Vậy\,x > \dfrac{9}{4}\,hoac:x < 0\\
14)\\
\left| {{x^2} - 7x + 2} \right| \le {x^2} - 9\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
- {x^2} + 9 \le {x^2} - 7x + 2 \le {x^2} - 9
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
2{x^2} - 7x - 7 \ge 0\\
7x \ge 11
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
\left[ \begin{array}{l}
x \ge \dfrac{{7 + \sqrt {105} }}{4}\\
x \le \dfrac{{7 - \sqrt {105} }}{4}
\end{array} \right.\\
x \ge \dfrac{{11}}{7}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge \dfrac{{7 + \sqrt {105} }}{4}\\
x \le - 3
\end{array} \right.
\end{array}$
