Cho $x=2018!$. Tính $A=\frac{1}{\log_{2^{2018}}x}+\frac{1}{\log_{3^{2018}}x}+...+\frac{1}{\log_{2017^{2018}}x}+\frac{1}{\log_{2018^{2018}}x}$.

Question

Beedemy · Accepted Answer

Lời giải: $A=\frac{1}{\log_{2^{2018}}x}+\frac{1}{\log_{3^{2018}}x}+...+\frac{1}{\log_{2017^{2018}}x}+\frac{1}{\log_{2018^{2018}}x}=\log_{x}2^{2018}+\log_{x}3^{2018}+...+\log_{x}2017^{2018}+\log_{x}2018^{2018}=2018.\log_{x}2+2018.\log_{x}3+...+2018.\log_{x}2017+2018.\log_{x}2018=2018.\left( \log_{x}2+\log_{x}3+...+\log_{x}2017+\log_{x}2018 \right)=2018.\log_{x}\left( 2.3.....2017.2018 \right)$

Bài tập nâng cao Lôgarit · Bài 18