Đáp án:
$\dfrac{1}{{a + 1}}\left( {1 + \dfrac{1}{b}} \right)$
Giải thích các bước giải:
$\begin{array}{l}
{\log _{15}}\left( {21} \right) = {\log _{15}}3 + {\log _{15}}7\\
= \dfrac{1}{{{{\log }_3}15}} + \dfrac{1}{{{{\log }_7}15}} = \dfrac{1}{{1 + {{\log }_3}5}} + \dfrac{1}{{{{\log }_7}3 + {{\log }_7}5}}\\
= \dfrac{1}{{1 + {{\log }_3}5}} + \dfrac{1}{{{{\log }_7}5.{{\log }_5}3 + {{\log }_7}5}}\\
= \dfrac{1}{{1 + a}} + \dfrac{1}{{ab + b}} = \dfrac{1}{{1 + a}} + \dfrac{1}{{b\left( {a + 1} \right)}} = \dfrac{1}{{a + 1}}\left( {1 + \dfrac{1}{b}} \right)
\end{array}$
