Đáp án:
$C. \dfrac{a+2ab}{ab+b}$
Giải thích các bước giải:
Câu 12:
$\log_6 45=\log_6 (3^2.5)=\log_6 3^2+ \log_6 5=2\log_6 3+\log_6 5$
$=\dfrac{2}{\log_3 6}+\dfrac{1}{\log_5 6}=\dfrac{2}{\log_3 (2.3)}+\dfrac{1}{\log_5 (2.3)}$
$=\dfrac{2}{1+\log_3 2}+\dfrac{1}{\log_5 2+\log_5 3}=\dfrac{2}{1+\dfrac{1}{\log_2 3}}+\dfrac{1}{\log_5 3.\log_3 2+\log_53}$
$=\dfrac{2}{1+\dfrac{1}{\log_2 3}}+\dfrac{1}{\log_5 3.\dfrac{1}{\log_2 3}+\log_5 3}=\dfrac{2}{1+\dfrac{1}{a}}+\dfrac{1}{b.\dfrac{1}{a}+b}$
$=\dfrac{2a}{a+1}+\dfrac{a}{ab+b}=\dfrac{2a}{a+1}+\dfrac{a}{b(a+1)}$
$=\dfrac{2ab+a}{b(a+1)}=\dfrac{a+2ab}{ab+b}$
$\to C$
