Lời giải:
\(\frac{2022\times 2023-3}{2023\times 2021+2020}=\frac{2023\times (2021+1)-3}{2023\times 2021+2020} \\ =\frac{2023\times 2021+2023-3}{2023\times 2021+2020}=\frac{2023\times 2021+2020}{2023\times 2021+2020}=1\)

A = 2019 \(\times\) 2021 + 2023

A = (2018 + 1).(2022 -1) + 2023 

A = 2018.2022 - 2018 + 2023 > 2018.2022 - 2022

Vậy A > B 

Cách 1: Nhìn qua là biết A > B :)) 

Cách 2: Giải cụ thể:

A = 2019 x 2021 + 2023

   = 2018 x 2021 + 2021 + 2023 = 2018 x 2021 + 4044

B = 2018 x 2022 - 2022

   = 2018 x 2021 + 2018 - 2022 = 2018 x 2021 - 4

⇒ A > B và lớn hơn: 4044 + 4 = 4048

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

\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)

= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)

= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)

= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)

= 1

2023×2021+20222022×2023−1​

= (2021+1)×2023−12023×2021+20222023×2021+2022(2021+1)×2023−1​

= 2023×2021+2023−12023×2021+20222023×2021+20222023×2021+2023−1​

= 2023×2021+20222023×2021+20222023×2021+20222023×2021+2022​

= 1

Không cần tính, ta thấy : 2022/2021 > 2021/2022
Vậy : 2022/2021*2023 > 2021/2022*2022

giúp vơi

khó difficult