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Câu hỏi: Cho a là số thực dương khác 1.
Biểu thức $P={{\log }_{a}}2019+{{\log }_{\sqrt{a}}}2019+{{\log }_{\sqrt[3]{a}}}2019+...+{{\log }_{\sqrt[2018]{a}}}2019+{{\log }_{\sqrt[2019]{a}}}2019$ bằng:
A. $1010.2019.{{\log }_{a}}2019.$
B. $2018.2019.{{\log }_{a}}2018.$
C. $2018.{{\log }_{a}}2018.$
D. $2019.{{\log }_{a}}2018.$
Biểu thức $P={{\log }_{a}}2019+{{\log }_{\sqrt{a}}}2019+{{\log }_{\sqrt[3]{a}}}2019+...+{{\log }_{\sqrt[2018]{a}}}2019+{{\log }_{\sqrt[2019]{a}}}2019$ bằng:
A. $1010.2019.{{\log }_{a}}2019.$
B. $2018.2019.{{\log }_{a}}2018.$
C. $2018.{{\log }_{a}}2018.$
D. $2019.{{\log }_{a}}2018.$
Ta có:
$\begin{aligned}
& P={{\log }_{a}}2019+{{\log }_{\sqrt{a}}}2019+{{\log }_{\sqrt[3]{a}}}2019+...+{{\log }_{\sqrt[2018]{a}}}2019+{{\log }_{\sqrt[2019]{a}}}2019 \\
& \ \ \ ={{\log }_{a}}2019+2.{{\log }_{a}}2019+3.{{\log }_{a}}2019+...+2019{{\log }_{a}}2019 \\
& \ \ \ =\left( 1+2+3+...+2019 \right).{{\log }_{a}}2019 \\
& \ \ \ =\dfrac{2019}{2}\left( 1+2019 \right).{{\log }_{a}}2019=1010.2019.{{\log }_{a}}2019. \\
\end{aligned}$
$\begin{aligned}
& P={{\log }_{a}}2019+{{\log }_{\sqrt{a}}}2019+{{\log }_{\sqrt[3]{a}}}2019+...+{{\log }_{\sqrt[2018]{a}}}2019+{{\log }_{\sqrt[2019]{a}}}2019 \\
& \ \ \ ={{\log }_{a}}2019+2.{{\log }_{a}}2019+3.{{\log }_{a}}2019+...+2019{{\log }_{a}}2019 \\
& \ \ \ =\left( 1+2+3+...+2019 \right).{{\log }_{a}}2019 \\
& \ \ \ =\dfrac{2019}{2}\left( 1+2019 \right).{{\log }_{a}}2019=1010.2019.{{\log }_{a}}2019. \\
\end{aligned}$
Đáp án A.