Đáp án:
$B$ > $A$
Giải thích các bước giải:
Ta có:
$A$ = $\frac{2020^{2018}-1 }{2020^{2019} - 2019}$
⇒ $2020A$ = $\frac{2020^{2019}-2020 }{2020^{2019} - 2019}$
⇒ $2020A$ = $1$ - $\frac{1}{2020^{2019} - 2019}$
$B$ = $\frac{2020^{2019} + 1}{2020^{2020} + 2019}$
⇒ $2020B$ = $\frac{2020^{2020} + 2020 }{2020^{2020} + 2019}$
⇒ $2020B$ = $1$ + $\frac{1}{2020^{2020}+2019}$
Do $2020B$ > $1$
$2020A$ < $1$
Nên $2020B$ > $2020A$
⇒ $B$ > $A$
Vậy $B$ > $A$