Ta có

B

- Hướng dẫn giải

Đáp án B

Ta có 

$\begin{array}{l}{\log }_{6}28={\log }_{6}3.{\log }_{3}28\\ =\frac{{\log }_{3}28}{{\log }_{3}6}=\frac{{\log }_3\left(7.4\right)}{{\log }_3\left(2.3\right)}\end{array}$

$\begin{array}{l}=\frac{{\log }_3{2}^2+{\log }_{3}7}{{\log }_{3}2+1}\\ =\frac{2{\log }_{3}2+{\log }_{3}7}{1+{\log }_{3}2}\\ =\frac{2{\log }_{3}2+2+{\log }_{3}7-2}{1+{\log }_{3}2}\\ =2+\frac{{\log }_{3}7-2}{{\log }_{3}2+1}\end{array}$

$\Rightarrow a+b+c=2-2+1=1$