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Câu hỏi: Tính $P={{\log }_{12}}6+{{\log }_{\sqrt{12}}}6+{{\log }_{\sqrt[3]{12}}}6+...+{{\log }_{\sqrt[2020]{12}}}6.$
A. $2020.{{\log }_{12}}6.$
B. $2020.{{\log }_{6}}12.$
C. $2041210.{{\log }_{12}}6.$
D. $2041210.{{\log }_{6}}12.$
A. $2020.{{\log }_{12}}6.$
B. $2020.{{\log }_{6}}12.$
C. $2041210.{{\log }_{12}}6.$
D. $2041210.{{\log }_{6}}12.$
Ta có $P={{\log }_{12}}6+2{{\log }_{12}}6+3{{\log }_{12}}6+...+2020{{\log }_{12}}6$
$=\left( 1+2+3+2020 \right).{{\log }_{12}}6=\dfrac{2020.2021}{2}.{{\log }_{12}}6=2041210.{{\log }_{12}}6$.
$=\left( 1+2+3+2020 \right).{{\log }_{12}}6=\dfrac{2020.2021}{2}.{{\log }_{12}}6=2041210.{{\log }_{12}}6$.
Đáp án C.