Tham khảo:

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

a ) <

b ) <

-2021/2020<1/2

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

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

Ta có:2019>4
=>2019/2020+2020/2021+2021/2022+2019>4
=>a>4(dpcm)

\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)

\(B=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)

Cứu với ;-;