@Sammi67852

Đáp án + Giải thích các bước giải:

$log_{27}5=log_{3^3}5=\frac{1}{3}log_{3}5=a$

⇒$log_{3}5=3a->log_{5}3=\frac{1}{3a}$ 

 $log_{8}7=log_{2^3}7=\frac{1}{3}log_{2}7=b$

⇒$log_{2}7=3b->log_{7}2=\frac{1}{3b}$

`log_{2}3=c->`log_{3}2=\frac{1/c}`

Có: `log_{6}35=log_{6}7+log_{6}5`

⇒  $\frac{log_{2}7}{log_{2}6}$ + $\frac{log_{3}5}{log_{3}6}$

$=\frac{log_{2}7}{log_{2}2+log_{2}3}$ + $\frac{log_{3}5}{log_{3}3+log_{3}2}$

$=\frac{3b}{1+c}$ + $\frac{3a}{1+\frac{1}{c}}$

$=\frac{3b}{1+c}$ + $\frac{3a}{\frac{c+1}{c}}$

$=\frac{3b}{1+c}$ + $\frac{3ac}{c+1}$=$\frac{3b+3ac}{c+1}$ 

⇒`m+n=3+1=4`