Đáp án:
9D, 10B
Giải thích các bước giải:
$\begin{array}{l}
9)\frac{1}{{{{\log }_2}n!}} + \frac{1}{{{{\log }_3}n!}} + ... + \frac{1}{{{{\log }_n}n!}}\\
= {\log _{n!}}2 + {\log _{n!}}3 + ... + {\log _{n!}}n\\
= {\log _{n!}}\left( {2.3.....n} \right) = {\log _{n!}}n! = 1\\
10)S = 1 + {2^2}{\log _{\sqrt 2 }}2 + {3^2}{\log _{\sqrt[3]{2}}}2 + ... + {2018^2}{\log _{\sqrt[{2018}]{2}}}2\\
S = 1 + {2^2}.{\log _{{2^{\frac{1}{2}}}}}2 + {3^2}{\log _{{2^{\frac{1}{3}}}}}2 + ... + {2018^2}{\log _{{2^{\frac{1}{{2018}}}}}}2\\
S = 1 + {2^2}.2 + {3^2}.3 + ... + {2018^2}.2018\\
S = 1 + {2^3} + {3^3} + ... + {2018^3}\\
S = {\left( {1 + 2 + 3 + ... + 2018} \right)^2}\\
S = {\left( {\frac{{2018.2019}}{2}} \right)^2}\\
S = {1009^2}{.2019^2}
\end{array}$
