Cho

B

- Hướng dẫn giải

Đáp án B

Ta có: $a={\log }_{12}27=3{\log }_{12}3=\frac{3}{{\log }_{3}12}=\frac{3}{2{\log }_{3}2+1}$

$\Rightarrow {\log }_{2}3=\frac{2a}{3-a}$ 

Vậy:

$T={\log }_{36}24=\frac{1}{2}\left({\log }_{6}4+1\right)=\frac{1}{2}+{\log }_{6}2=\frac{1}{2}+\frac{1}{{\log }_{2}6}$ 

 $=\frac{1}{2}+\frac{1}{1+{\log }_{2}3}=\frac{1}{2}+\frac{1}{1+\frac{2a}{3-a}}=\frac{1}{2}+\frac{3-a}{3+a}=\frac{9-a}{6+a}$ .