Cho $\Large \mathrm{log}_35=a$, $\Large \mathrm{log}_52=b$, $\Large \m

Cho $\Large \mathrm{log}_35=a$, $\Large \mathrm{log}_52=b$, $\Large \mathrm{log}_311=c$. Khi đó $\Large \mathrm{log}_{216}495$ bằng

• A.

  $\Large \dfrac{a+c+2}{3ab+3}$.

• B.

  $\Large \dfrac{a+c+2}{3ab}$.

• C.

  $\Large \dfrac{a+c}{3ab+3}$.

• D.

  $\Large \dfrac{a+c+2}{ab+3}$.

Chọn A

Cách 1

$\Large \mathrm{log}_{216}495=\mathrm{log}_{2^3.3^3}(3^2.5.11)$ $\Large =\mathrm{log}_{2^3.3^3}3^2+\mathrm{log}_{2^3.3^3}5+\mathrm{log}_{2^3.3^3}11$

$\Large =\dfrac{1}{\mathrm{log}_{3^2}2^3+\mathrm{log}_{3^2}3^3}+\dfrac{1}{\mathrm{log}_52^3+\mathrm{log}_53^3}+\dfrac{1}{\mathrm{log}_{11}2^3+\mathrm{log}_{11}3^3}$

$\Large =\dfrac{1}{\dfrac{3}{2}\mathrm{log}_32+\dfrac{3}{2}}+\dfrac{1}{3\mathrm{log}_52+3\mathrm{log}_53}+\dfrac{1}{3\mathrm{log}_{11}2+3\mathrm{log}_{11}3}$

$\Large =\dfrac{1}{\dfrac{3}{2}.\dfrac{\mathrm{log}_52}{\mathrm{log}_53}+\dfrac{3}{2}}+\dfrac{1}{3b+\dfrac{3}{a}}+\dfrac{1}{3.\dfrac{\mathrm{log}_32}{\mathrm{log}_311}+\dfrac{3}{c}}$

$\Large =\dfrac{1}{\dfrac{3}{2}.ab+\dfrac{3}{2}}+\dfrac{1}{3b+\dfrac{3}{a}}+\dfrac{1}{3.\dfrac{ab}{c}+\dfrac{3}{c}}$ $\Large =\dfrac{2}{3ab+3}+\dfrac{a}{3ab+3}+\dfrac{c}{3ab+3}=\dfrac{a+c+2}{3ab+3}$.