Đáp án:
$2021^{2020} > 2020^{2019} + 2020^{2020}$
Giải thích các bước giải:
$+) \quad 2020^{2019} + 2020^{2020}$
$= 2020^{2019} + 2020^{2019}.2020$
$= 2020^{2019}.2021$
$+) \quad 2021^{2020}$
$= 2021^{2019}.2021$
Do $2021 > 2020$
$\to 2021^{2019} > 2020^{2019}$
$\to 2021^{2019}.2021 > 2020^{2019}.2021$
$\to 2021^{2020} > 2020^{2019} + 2020^{2020}$