Biết ${{\log }_{15}}20=a+\dfrac{2{{\log }_{3}}2+b}{{{\log }_{3}}5+c}$ với $a, b, c\in \mathbb{Z}$. Tính $T=a+b+c$.

${{\log }_{15}}20=\dfrac{{{\log }_{3}}20}{{{\log }_{3}}15}=\dfrac{{{\log }_{3}}\left( {{2}^{2}}.5 \right)}{{{\log }_{3}}\left( 5.3 \right)}=\dfrac{2{{\log }_{3}}2+{{\log }_{3}}5}{{{\log }_{3}}5+1}$ $=\dfrac{2{{\log }_{3}}2-1+1+{{\log }_{3}}5}{{{\log }_{3}}5+1}=1+\dfrac{2{{\log }_{3}}2-1}{{{\log }_{3}}5+1}$.
Do đó $a=1$ ; $b=-1$ ; $c=1\Rightarrow T=a+b+c=1$.