a) Ta có
$\sqrt{2020} - \sqrt{2021} = \dfrac{2020- 2021}{\sqrt{2020} + \sqrt{2021}} = -\dfrac{1}{\sqrt{2020} + \sqrt{2021}}$
Lại có
$\sqrt{2019} - \sqrt{2020} = \dfrac{2019 - 2020}{\sqrt{2019} + \sqrt{2020}} = -\dfrac{1}{\sqrt{2019} + \sqrt{2020}$
Ta có
$\sqrt{2019} + \sqrt{2020} < \sqrt{2020} + \sqrt{2021}$
$<-> \dfrac{1}{\sqrt{2019} + \sqrt{2020}} > \dfrac{1}{\sqrt{2020} + \sqrt{2021}}$
$<-> -\dfrac{1}{\sqrt{2019} + \sqrt{2020}} < - \dfrac{1}{\sqrt{2020} + \sqrt{2021}}$
$<-> \sqrt{2019} - \sqrt{2020} < \sqrt{2020} - \sqrt{2021}$
Vậy $\sqrt{2019} - \sqrt{2020} < \sqrt{2020} - \sqrt{2021}$.
b) Ta có
$(\sqrt{2019} + \sqrt{2021})^2 = 2019 + 2021 + 2 \sqrt{2019.2021} = 4040 + 2\sqrt{2019 . 2021}$
và
$(2\sqrt{2020})^2 = 4 . 2020 = 4040 . 2$
Khi đó ta có
$(\sqrt{2019} + \sqrt{2021})^2 - (2\sqrt{2020})^2 = 2\sqrt{2019 . 2021} - 4040$
$= 2(\sqrt{2019 . 2021} - 2020)$
Ta sẽ so sánh $2019 . 2021$ và $2020^2$
Ta có
$2019 . 2021 = (2020 - 1)(2020 + 1) = 2020^2 - 1 < 2020^2$
Suy ra
$\sqrt{2019 . 2021} - 2020 < 0$
$<-> (\sqrt{2019} + \sqrt{2021})^2 - (2\sqrt{2020})^2 <0$
$<-> \sqrt{2019} + \sqrt{2021} < 2\sqrt{2020}$
