Ta có : \(\sqrt{2021}-\sqrt{2020}\) = \(\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
\(\sqrt{2019}-\sqrt{2018}=\frac{2019-2018}{\sqrt{2019}+\sqrt{2018}}=\frac{1}{\sqrt{2019}+\sqrt{2018}}\)
Dễ thấy : \(2021+2020>2019+2018\)
=> \(\sqrt{2021}+\sqrt{2020}>\sqrt{2019}+\sqrt{2018}\)
từ đây suy ra : \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2019}+\sqrt{2018}}\) ( theo so sánh các phân số cùng tử số )
Vậy \(\sqrt{2021}-\sqrt{2020}< \sqrt{2019}-\sqrt{2018}\)
Ta có:\(\left(\sqrt{2021}+\sqrt{2018}\right)^2=2021+2018+2\sqrt{2021.2018}\)
Mà 2021.2018=2020.2018+2018<2020.2018+2020=2020.2019
\(\Rightarrow2021+2018+2\sqrt{2021.2018}< 2021+2018+2\sqrt{2019.2020}=2019+2\sqrt{2019.2020}+2020=\left(\sqrt{2019}+\sqrt{2020}\right)^2\)
\(\Rightarrow\left(\sqrt{2021}+\sqrt{2018}\right)^2< \left(\sqrt{2020}+\sqrt{2019}\right)^2\)
Vì cả 2 vế đều lớn hơn 0 nên
\(\Rightarrow\sqrt{2021}+\sqrt{2018}< \sqrt{2020}+\sqrt{2019}\Rightarrow\sqrt{2021}-\sqrt{2020}< \sqrt{2019}-\sqrt{2018}\)
