$a={{\log }_{27}}5={{\log }_{{{3}^{3}}}}5=\frac{1}{3}{{\log }_{3}}5\Rightarrow {{\log }_{3}}5=3a$

$b={{\log }_{8}}7={{\log }_{{{2}^{3}}}}7=\frac{1}{3}{{\log }_{2}}7\Rightarrow {{\log }_{2}}7=3b$

Khi đó ${{\log }_{12}}35={{\log }_{12}}5+{{\log }_{12}}7=\frac{1}{{{\log }_{5}}12}+\frac{1}{{{\log }_{7}}12}=\frac{1}{{{\log }_{5}}3+{{\log }_{5}}4}+\frac{1}{{{\log }_{7}}3+{{\log }_{7}}4}$

Ta có $c.3a={{\log }_{2}}3.{{\log }_{3}}5={{\log }_{2}}5\Rightarrow {{\log }_{5}}4=2{{\log }_{5}}2=\frac{2}{{{\log }_{2}}5}=\frac{2}{3ac}$

$\frac{1}{3b}.c={{\log }_{7}}2.{{\log }_{2}}3={{\log }_{7}}3;{{\log }_{7}}4=2{{\log }_{7}}2=\frac{2}{{{\log }_{2}}7}=\frac{2}{3b}$

$\Rightarrow {{\log }_{12}}35=\frac{1}{\frac{1}{3a}+\frac{2}{3ac}}+\frac{1}{\frac{c}{3b}+\frac{2}{3b}}=\frac{3ac}{c+2}+\frac{3b}{c+2}=\frac{3b+3ac}{c+2}$