Đáp án: A
Giải thích các bước giải:
$\begin{array}{l}
{\log _{98}}32 = \dfrac{1}{{{{\log }_{32}}98}}\\
= \dfrac{1}{{{{\log }_{{2^5}}}\left( {14.7} \right)}}\\
= \dfrac{1}{{\dfrac{1}{5}.{{\log }_2}\left( {14.7} \right)}}\\
= \dfrac{5}{{{{\log }_2}14 + {{\log }_2}7}}\\
= \dfrac{5}{{x + {{\log }_2}\dfrac{{14}}{2}}}\\
= \dfrac{5}{{x + {{\log }_2}14 - {{\log }_2}2}}\\
= \dfrac{5}{{x + x - 1}}\\
= \dfrac{5}{{2x - 1}} = \dfrac{a}{{b.x - c}}\\
\Rightarrow \left\{ \begin{array}{l}
a = 5\\
b = 2\\
c = 1
\end{array} \right.\\
\Rightarrow S = 2a + 3b + 5c\\
= 2.5 + 3.2 + 5.1\\
= 21
\end{array}$
