Question

Answer 1

giúp mình nha

Answer 2

a: \(10=7+3\)

mà \(3>2\sqrt{2}\)

nên \(7+2\sqrt{2}< 10\)

Answer 3

Lời giải:
a. $7+2\sqrt{2}=7+\sqrt{8}<7+\sqrt{9}=7+3=10$
b.

$\sqrt{2021}-\sqrt{2019}=\frac{2021-2019}{\sqrt{2021}+\sqrt{2019}}=\frac{2}{\sqrt{2021}+\sqrt{2019}}$

$\sqrt{2018}-\sqrt{2016}=\frac{2018-2016}{\sqrt{2018}+\sqrt{2016}}$

$=\frac{2}{\sqrt{2018}+\sqrt{2016}}$

Dễ thấy $\frac{2}{\sqrt{2021}+\sqrt{2019}}< \frac{2}{\sqrt{2018}+\sqrt{2016}}$ do $\sqrt{2021}+\sqrt{2019}> \sqrt{2018}+\sqrt{2016}$

Do đó:

$\sqrt{2021}-\sqrt{2019}< \sqrt{2018}-\sqrt{2016}$