Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
9,\\
- {x^2} + 6x - 5 \le 0\\
\Leftrightarrow {x^2} - 6x + 5 \ge 0\\
\Leftrightarrow \left( {{x^2} - x} \right) - \left( {5x - 5} \right) \ge 0\\
\Leftrightarrow x\left( {x - 1} \right) - 5\left( {x - 1} \right) \ge 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 5} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 5\\
x \le 1
\end{array} \right.\\
\Rightarrow S = \left( { - \infty ;1} \right] \cup \left[ {5; + \infty } \right)\\
12,\\
a,\\
3x - \frac{5}{{{x^2} + 1}} > 1 - \frac{5}{{{x^2} + 1}}\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 1 \ne 0\\
3x > 1
\end{array} \right. \Leftrightarrow 3x - 1 > 0\,\,\,\,\,\left( {t/m} \right)\\
b,\\
3x + \frac{1}{{x - 2}} > \frac{1}{{x - 2}} + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ne 0\\
3x > 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x - 1 > 0\\
x \ne 2
\end{array} \right.\,\,\,\,\,\left( L \right)\\
c,\\
9{x^2} - 1 > 0 \Leftrightarrow \left( {3x - 1} \right)\left( {3x + 1} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
3x - 1 > 0\\
3x + 1 < 0
\end{array} \right.\,\,\,\,\,\left( L \right)\\
d,\\
- 6x + 2 > 0 \Leftrightarrow - 2\left( {3x - 1} \right) > 0 \Leftrightarrow 3x - 1 < 0\,\,\,\,\,\,\,\left( L \right)
\end{array}\)
