Đáp án:
$\begin{array}{l}
18){\log _{0,3}}\left( {4{x^2}} \right) \ge {\log _{0,3}}\left( {12x - 5} \right)\left( {dk:x \ne 0;x > \frac{5}{{12}}} \right)\\
\Leftrightarrow 4{x^2} \le 12x - 5\\
\Leftrightarrow 4{x^2} - 12x + 5 \le 0\\
\Leftrightarrow \left( {x - \frac{5}{2}} \right)\left( {x - \frac{1}{2}} \right) \le 0\\
\Leftrightarrow \frac{1}{2} \le x \le \frac{5}{2}\\
\Rightarrow M = \frac{5}{2};m = \frac{1}{2} \Rightarrow M + m = 3 \Rightarrow A\\
19)Đk:\left\{ \begin{array}{l}
x > 0\\
1 + {\log _{\frac{1}{9}}}x - {\log _9}x > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x > 0\\
1 - 2{\log _9}x > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 0\\
{\log _9}x < \frac{1}{2}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x > 0\\
x < \sqrt 9
\end{array} \right. \Rightarrow 0 < x < 3\\
Ta\,có:{\log _2}\left( {1 + {{\log }_{\frac{1}{9}}}x - {{\log }_9}x} \right) < 1\\
\Rightarrow 1 - {\log _9}x - {\log _9}x < 2\\
\Rightarrow {\log _9}x > \frac{{ - 1}}{2}\\
\Rightarrow x > {9^{ - 1/2}}\\
\Rightarrow x > \frac{1}{{\sqrt 9 }} \Rightarrow x > \frac{1}{3}\\
Vay\,\left( {\frac{1}{a};b} \right) = \left( {\frac{1}{3};3} \right) \Rightarrow a = b = 3 \Rightarrow C
\end{array}$