so sánh A=2018 ^2018 +1 / 2018 ^2019 +1

* Hướng dẫn giải

Đáp án cần chọn là: A

Dễ thấy A < 1 nên:

$\begin{array}{l}A=\frac{{2018}^{2018}+1}{{2018}^{2019}+1}<\frac{\left({2018}^{2018}+1\right)+2017}{\left({2018}^{2019}+1\right)+2017}\\ =\frac{{2018}^{2018}+2018}{{2018}^{2019}+2018}=\frac{2018.\left({2018}^{2017}+1\right)}{2018.\left({2018}^{2018}+1\right)}\\ =\frac{{2018}^{2017}+1}{{2018}^{2018}+1}=B\end{array}$

Vậy A < B