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