Cho

A

- Hướng dẫn giải

Đáp án A

Cách 1: Tư duy tự luận

$\begin{array}{l}{\log }_{27}5={\log }_{3^3}5=\frac{1}{3}{\log }_{3}5=a\\ \Rightarrow {\log }_{3}5=3a\Rightarrow {\log }_{5}3=\frac{1}{3a}\end{array}$.

$\begin{array}{l}{\log }_{5}4={\log }_{5}3.{\log }_{3}2.{\log }_{2}4\\ =\frac{2{\log }_{5}3}{{\log }_{2}3}=\frac{2}{3ac}\end{array}$ .

$\begin{array}{l}{\log }_{8}7={\log }_{2^3}7\\ =\frac{1}{3}{\log }_{2}7=b\\ \Rightarrow {\log }_{2}7=3b\\ \Rightarrow {\log }_{7}2=\frac{1}{3b}\\ \Rightarrow {\log }_{7}4=2{\log }_{7}2=\frac{2}{3b}.\end{array}$

${\log }_{7}3={\log }_{7}2.{\log }_{2}3=\frac{c}{3b}$.

$\begin{array}{l}{\log }_{12}35={\log }_{12}5+{\log }_{12}7\\ =\frac{1}{{\log }_{5}12}+\frac{1}{{\log }_{7}12}\\ =\frac{1}{{\log }_{5}3+{\log }_{5}4}+\frac{1}{{\log }_{7}3+{\log }_{7}4}\end{array}$ 

$\begin{array}{l}\Rightarrow {\log }_{12}35=\frac{1}{\frac{1}{3a}+\frac{2}{3ac}}+\frac{1}{\frac{c}{3b}+\frac{2}{3b}}\\ =\frac{3ac}{c+2}+\frac{3b}{c+2}=\frac{3b+3ac}{c+2}\end{array}$.

Cách 2: Sử dụng máy tính cầm tay