Tính P = log2(2018^4) - 1/1009 + ln(e^2018).
Giải thích
Chọn D
\(\begin{array}{l}P = {\log _{{2^{2018}}}}4 - \frac{1}{{1009}} + \ln {e^{2018}} = {\log _{{2^{2018}}}}{2^2} - \frac{1}{{1009}} + 2018\ln e\\P = \frac{2}{{2018}}{\log _2}2 - \frac{1}{{1009}} + 2018 = \frac{2}{{2018}} - \frac{1}{{1009}} + 2018 = 2018\end{array}\)