A = B

Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).

:D

Tham khảo:

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

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

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)

a: Đặt \(A=\sqrt{2021}-\sqrt{2020}\) và \(B=\sqrt{2022}-\sqrt{2021}\)

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

\(B=\sqrt{2022}-\sqrt{2021}=\frac{2022-2021}{\sqrt{2022}+\sqrt{2021}}=\frac{1}{\sqrt{2022}+\sqrt{2021}}\)

Ta có: \(\sqrt{2021}+\sqrt{2020}<\sqrt{2022}+\sqrt{2021}\)

=>\(\frac{1}{\sqrt{2021}+\sqrt{2020}}>\frac{1}{\sqrt{2022}+\sqrt{2021}}\)

=>A>B

b: Đặt \(A=\sqrt{2022}-\sqrt{2020}\) và \(B=\sqrt{2020}-\sqrt{2018}\)

\(A=\sqrt{2022}-\sqrt{2020}=\frac{2022-2020}{\sqrt{2022}+\sqrt{2020}}=\frac{2}{\sqrt{2022}+\sqrt{2020}}\)

\(B=\sqrt{2020}-\sqrt{2018}=\frac{2020-2018}{\sqrt{2020}+\sqrt{2018}}=\frac{2}{\sqrt{2020}+\sqrt{2018}}\)

TA có: \(\sqrt{2022}+\sqrt{2020}>\sqrt{2020}+\sqrt{2018}\)

=>\(\frac{2}{\sqrt{2022}+\sqrt{2020}}<\frac{2}{\sqrt{2020}+\sqrt{2018}}\)

=>A<B

Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)