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

Ta có:

\[lo{g_{15}}20 = lo{g_{15}}({2^2}.5)\]

\[ = 2lo{g_{15}}2 + lo{g_{15}}5\]

\[ = \frac{2}{{lo{g_2}15}} + \frac{1}{{lo{g_5}15}}\]

\[ = \frac{2}{{lo{g_2}3 + lo{g_2}5}} + \frac{1}{{lo{g_5}3 + lo{g_5}5}}\]

\[ = \frac{2}{{\frac{1}{{lo{g_3}2}} + \frac{{lo{g_3}5}}{{lo{g_3}2}}}} + \frac{1}{{lo{g_5}3 + 1}}\]

\[ = \frac{{2lo{g_3}2}}{{1 + lo{g_3}5}} + \frac{1}{{\frac{1}{{lo{g_3}5}} + 1}}\]

\[ = \frac{{2lo{g_3}2}}{{1 + lo{g_3}5}} + \frac{{lo{g_3}5}}{{lo{g_3}5 + 1}}\]

\[ = \frac{{2lo{g_3}2 + lo{g_3}5}}{{lo{g_3}5 + 1}}\]

\[ = \frac{{lo{g_3}5 + 1 + 2lo{g_3}2 - 1}}{{lo{g_3}5 + 1}}\]

\[ = 1 + \frac{{2lo{g_3}2 - 1}}{{lo{g_3}5 + 1}}\]

\[ \Rightarrow a = 1,\,\,b = - 1,\,\,c = 1\]

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

Đáp án cần chọn là: D