Đáp án: B
Giải thích các bước giải:
Do $a,b>0$
$\rightarrow \left\{ \begin{array}{l} 3a + 2b + 1 > 1\\ 9{a^2} + {b^2} + 1 > 1\\ 6ab + 1 > 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) > 0\\ {\log _{6ab + 1}}\left( {3a + 2b + 1} \right) > 0 \end{array} \right.$
$\rightarrow{\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) + {\log _{6ab + 1}}\left( {3a + 2b + 1} \right) \ge 2\sqrt {{{\log }_{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) + {{\log }_{6ab + 1}}\left( {3a + 2b + 1} \right)}$
$ \begin{array}{l} \Leftrightarrow 2 \ge 2\sqrt {{{\log }_{6ab + 1}}\left( {9{a^2} + {b^2} + 1} \right)} \Leftrightarrow {\log _{6ab + 1}}\left( {9{a^2} + {b^2} + 1} \right) \le 1 \Leftrightarrow 9{a^2} + {b^2} + 1 \le 6ab + 1\\ \Leftrightarrow {\left( {3a - b} \right)^2} \le 0 \Leftrightarrow 3a = b \end{array}$
$\rightarrow {\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) = {\log _{6ab + 1}}\left( {3a + 2b + 1} \right) \Leftrightarrow {\log _{3b + 1}}\left( {2{b^2} + 1} \right) = {\log _{2{b^2} + 1}}\left( {3b + 1} \right)$
$\rightarrow 2{b^2} + 1 = 3b + 1 \Leftrightarrow 2{b^2} - 3b = 0 \Leftrightarrow b = \frac{3}{2}$
$\rightarrow a=\dfrac{1}{2}$
$\rightarrow b=3a=\dfrac{3}{2}\rightarrow a+2b=\dfrac{7}{2}$
