Question

[Ngọc Hà]

So sánh

A. √2021 - √2020 và √2020 - √2019

B. √2019×2021 và 2020

C. √2019 + √2021 và 2√2020

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\sqrt{2021}-\sqrt{2020}\)

\(=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)

\(=\frac{1}{\sqrt{2020}+\sqrt{2021}}\)

Ta có: \(\sqrt{2020}-\sqrt{2019}\)

\(=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)

\(=\frac{1}{\sqrt{2019}+\sqrt{2020}}\)

Ta có: \(\sqrt{2020}+\sqrt{2021}>\sqrt{2019}+\sqrt{2020}\)

\(\Leftrightarrow\frac{1}{\sqrt{2020}+\sqrt{2021}}< \frac{1}{\sqrt{2019}+\sqrt{2020}}\)

hay \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)

b) Ta có: \(\sqrt{2019\cdot2021}\)

\(=\sqrt{\left(2020-1\right)\left(2020+1\right)}\)

\(=\sqrt{2020^2-1}\)

Ta có: \(2020=\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

nên \(\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow\sqrt{2019\cdot2021}< 2020\)

c) Ta có: \(\left(\sqrt{2019}+\sqrt{2021}\right)^2\)

\(=2019+2021+2\cdot\sqrt{2019\cdot2021}\)

\(=4040+2\sqrt{2019\cdot2021}\)

\(=4040+2\cdot\sqrt{2020^2-1}\)

Ta có: \(\left(2\sqrt{2020}\right)^2\)

\(=4\cdot2020\)

\(=4040+2\cdot2020\)

\(=4040+2\cdot\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)

\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)