A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) ;  B = \(\dfrac{2020+2021}{2021+2022}\)

B = \(\dfrac{2020+2021}{2021+2022}\)   = \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\)

\(\dfrac{2020}{2021}\)   > \(\dfrac{2020}{2021+2022}\)

\(\dfrac{2021}{2022}\)     > \(\dfrac{2021}{2021+2022}\)

Cộng vế với vế ta có:

A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) > \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\) = B

Vậy A > B

A =  \(\dfrac{10^{10}-1}{10^{11}-1}\) 

A \(\times\) 10 = \(\dfrac{(10^{10}-1)\times10}{10^{11}-1}\) = \(\dfrac{10^{11}-10}{10^{11}-1}\) = 1 - \(\dfrac{9}{10^{11}-1}\) < 1

B = \(\dfrac{10^{10}+1}{10^{11}+1}\)

B \(\times\) 10 = \(\dfrac{(10^{10}+1)\times10}{10^{11}+1}\)  = \(\dfrac{10^{11}+10}{10^{11}+1}\) = 1 + \(\dfrac{9}{10^{11}+1}\) > 1

Vì 10 A< 1< 10B

Vậy A < B

Tham khảo:

\(\sqrt{2021}-\sqrt{2020}=\dfrac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\) là nghịch đảo của \(\sqrt{2021}+\sqrt{2020}\) (đpcm)

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

\(=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\)(đpcm)

Nhỏ hơn

Ta có 2020/2021 <1

         2021/2022 <1

         2022/2023 <1

         2023/2024 <1

Suy ra A=(2021/2021+2021/2022 +2022/2023 +2023/2024) < (1+1+1+1)= 4

      Vậy A <4

Chúc bạn học tốt

\(\dfrac{2020}{2021}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2023}{2024}< 1\)

Do đó: A<4

\(2021\left(2020+2022\right)-2020\left(2021+2022\right)\\ =2021.2020+2021.2022-2020.2021-2020.2022\\ =\left(2021.2020-2020.2021\right)+\left(2021.2022-2020.2022\right)\\ =0+2022.\left(2021-2020\right)\\ =0+2022.1=2022\)

2020/2021<1

2021/2022<1

2022/2023<1

2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023

=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4

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