Cho a, b, c

C

- Hướng dẫn giải

Đáp án C

Đặt 2^a = 3^b = 18^c = t

=> a = log₂t, b = log₃t, c = log₁₈t

$$\begin{gathered} \Rightarrow T=\frac{b}{c}-\frac{b}{a} \\ =\frac{{\log }_{3}t}{{\log }_{18}t}-\frac{{\log }_{3}t}{{\log }_{2}t} \\ =\frac{{\log }_{3}t}{{\displaystyle \frac{{\log }_{3}t}{{\log }_{3}18}}}-\frac{{\log }_{3}t}{{\displaystyle \frac{{\log }_{3}t}{{\log }_{3}2}}} \\ ={\log }_{3}18-{\log }_{3}2 \\ ={\log }_{3}9=2 \end{gathered}$$