Ta có:
\(A=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
mà 2019+2020 >2019>2020 \(\Rightarrow\frac{2018}{2019+2020}< \frac{2018}{2019};\frac{2019}{2019+2020}< \frac{2019}{2020}\)
\(\Rightarrow\frac{2018}{2019+2020}+\frac{2019}{2019+2020}< \frac{2018}{2019}+\frac{2019}{2020}\)hay \(A< B\)
A = 2018+2019/2019 + 2020
= \(\frac{2019+2020-2}{2019+2020}\) = 1 - \(\frac{2}{2019+2020}\)< 1
Có 2018/2019 > 1/2
2019/2020 >1/2
=> B = 2018/2019 + 2019/2020 > 1/2 + 1/2 = 1.
Vậy B > A