Google
Câu hỏi: Tính $P=\dfrac{1}{{{\log }_{2}}2020!}+\dfrac{1}{{{\log }_{3}}2020!}+\dfrac{1}{{{\log }_{4}}2020!}+....+\dfrac{1}{{{\log }_{2020}}2020!}.$
A. $P=2020.$
B. $P=2020!.$
C. $P=\dfrac{1}{2020}.$
D. $P=1.$
A. $P=2020.$
B. $P=2020!.$
C. $P=\dfrac{1}{2020}.$
D. $P=1.$
Ta có $P={{\log }_{2020!}}2+{{\log }_{2020!}}3+{{\log }_{2020!}}4+...+{{\log }_{2020!}}2020$
$={{\log }_{2020!}}\left( 2.3.4...2020 \right)={{\log }_{2020!}}\left( 2020! \right)=1$.
$={{\log }_{2020!}}\left( 2.3.4...2020 \right)={{\log }_{2020!}}\left( 2020! \right)=1$.
Đáp án B.