A=$\frac{1}{1.2}$ +$\frac{1}{2.3}$+ $\frac{1}{3.4}$ ...$\frac{1}{2019.2020}$
=1-$\frac{1}{2}$+ $\frac{1}{2}$ -$\frac{1}{3}$ +$\frac{1}{3}$ -$\frac{1}{4}$ +...+$\frac{1}{2019}$ -$\frac{1}{2020}$
=1-$\frac{1}{2020}$
=$\frac{2019}{2020}$
Vì $\frac{1}{2^{2} }$<$\frac{1}{1.2}$
$\frac{1}{3^{2}}$ <$\frac{1}{2.3}$
$\frac{1}{4^{2}}$ <$\frac{1}{3.4}$
.....
$\frac{1}{2020^{2}}$ <$\frac{1}{2019.2020}$
⇔A>B
Vậy A>B
