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

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

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 

Tham khảo:

yynbjkgyg

x= 2002/3000

ko bt đúng ko mong bn nhắc nhở

thank you

  2017/2020<2019/2020<  1
   1< 2022/2021< 2023/2021
vậy phân số lớn nhất là 2023/2021

ta so sánh với 1:

 2017/2020<2019/2020<  1
   1< 2022/2021< 2023/2021
nên phân số lớn nhất là phân số cuối: 2023/2021

\(=2023-1^{2020}+1=2023\)

Ko tính đc bạn

X=-200

X=-2022 nhà lúc nãy mik nhầm mong bạn thông cảm 

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