B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

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

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)

\(\dfrac{154}{155}>\dfrac{154}{155+156}\)

\(\dfrac{155}{156}>\dfrac{155}{155+156}\)

=>154/155+155/156>(154+155)/(155+156)

=>A>B

b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)

2021/2022>2021/6069

2022/2023>2022/2069

2023/2024>2023/6069

=>D>C

Tham khảo:

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

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

Đây dang lớp mấy vây

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.