1 câu trả lời
Đặt: `\sqrt{2021} - \sqrt{2020} = a`
`\sqrt{2022} - \sqrt{2021} = b`
`⇒ a - b = \sqrt{2021} - \sqrt{2020} - (\sqrt{2022} - \sqrt{2021})`
`= \sqrt{2021} - \sqrt{2020} - \sqrt{2022} + \sqrt{2021})`
`= 2\sqrt{2021} - (\sqrt{2020} + \sqrt{2022})`
$\\$
Ta lại có:
$\bullet$ `(\sqrt{2020} + \sqrt{2022})^2`
`= 2020 + 2022 + 2\sqrt{2020.2022}`
`= 4042 + 2\sqrt{(2021 - 1).(2021 + 1)}`
`= 4042 + 2\sqrt{2021^2 - 1}`
$\bullet$ `(2\sqrt{2021})^2 = 4.2021 = 4042 + 2.2021 = 4042 + 2\sqrt{2021^2}`
Vì: `2\sqrt{2021^2 - 1} < 2\sqrt{2021^2}`
`⇒ 4042 + 2\sqrt{2021^2 - 1} < 4042 + 2\sqrt{2021^2}`
Hay: `(\sqrt{2020} + \sqrt{2022})^2 < (2\sqrt{2021})^2`
`⇒ \sqrt{2020} + \sqrt{2022} < 2\sqrt{2021}`
`⇔ 2\sqrt{2021} - (\sqrt{2020} + \sqrt{2022}) > 0`
`⇒ a - b > 0`
`⇔ a > b`
Vậy: `\sqrt{2021} - \sqrt{2020} > \sqrt{2022} - \sqrt{2021}`