Biết \({\log _{15}}20 = a + \dfrac{{2{{\log }_3}2 + b}}{{{{\log }_3}5 + c}}\) với \(a,\,\,b,\,\,c \in \mathbb{Z}\). Tính \(T = a + b + c\).

Ta có:

\(\begin{array}{l}{\log _{15}}20 = {\log _{15}}\left( {{2^2}.5} \right)\\ = 2{\log _{15}}2 + {\log _{15}}5\\ = \dfrac{2}{{{{\log }_2}15}} + \dfrac{1}{{{{\log }_5}15}}\\ = \dfrac{2}{{{{\log }_2}3 + {{\log }_2}5}} + \dfrac{1}{{{{\log }_5}3 + {{\log }_5}5}}\\ = \dfrac{2}{{\dfrac{1}{{{{\log }_3}2}} + \dfrac{{{{\log }_3}5}}{{{{\log }_3}2}}}} + \dfrac{1}{{{{\log }_5}3 + 1}}\\ = \dfrac{{2{{\log }_3}2}}{{1 + {{\log }_3}5}} + \dfrac{1}{{\dfrac{1}{{{{\log }_3}5}} + 1}}\\ = \dfrac{{2{{\log }_3}2}}{{1 + {{\log }_3}5}} + \dfrac{{{{\log }_3}5}}{{{{\log }_3}5 + 1}}\\ = \dfrac{{2{{\log }_3}2 + {{\log }_3}5}}{{{{\log }_3}5 + 1}}\\ = \dfrac{{{{\log }_3}5 + 1 + 2{{\log }_3}2 - 1}}{{{{\log }_3}5 + 1}}\\ = 1 + \dfrac{{2{{\log }_3}2 - 1}}{{{{\log }_3}5 + 1}}\end{array}\)

\( \Rightarrow a = 1,\,\,b =  - 1,\,\,c = 1\).

Vậy \(T = a + b + c = 1 + \left( { - 1} \right) + 1 = 1.\)