Đặt $A = \dfrac{2020^{2019}+1}{2020^{2020}-1}$
$\to 2020A = \dfrac{2020^{2020}+2020}{2020^{2020}-1}$
$ = 1+ \dfrac{2021}{2020^{2020}-1}$
$B = \dfrac{2020^{2018}+1}{2020^{2019}-1}$
$\to 2020.B = \dfrac{2020^{2019}+2020}{2020^{2019}-1}$
$ = 1+\dfrac{2021}{2020^{2019}-1}$
Vì $\dfrac{2021}{2020^{2020}-1} < \dfrac{2021}{2020^{2019}-1}$
$\to 1+\dfrac{2021}{2020^{2020}-1} < 1+\dfrac{2021}{2020^{2019}-1}$
Hay $A<B$