So sánh: 2020^2019 + 2020^2020 và 2021^2020.

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1. `2020^2019+2020^2020=2020^2019 . (1+2020)=2021 . 2020^2019`

   `2021^2020=2021 . 2021^2019`

   Do `0<2020<2021`

   `=> 2021 . 2020^2019<2021 . 2021^2019`

   `=> 2020^2019+2020^2020<2021^2020`

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2. 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 $2021 > 2020$

   $\to 2021^{2019} > 2020^{2019}$

   $\to 2021^{2019}.2021 > 2020^{2019}.2021$

   $\to 2021^{2020} > 2020^{2019} + 2020^{2020}$

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