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

a 

oki

giúp vơi

khó difficult

Tham khảo:

A = 2021/2022+2020/2021+2019/2020+2018/2019+2017/2018

A<2022/2022+2021/2021+2020/2020+2019/2019+2018/2018

A<1+1+1+1+1

A<5

bn tham khảo link này nhé

https://olm.vn/hoi-dap/tim-kiem?id=1300742&subject=1&q=++++++++++kh%C3%B4ng+t%C3%ADnh+k%E1%BA%BFt+qu%E1%BA%A3+c%E1%BB%A5+th%E1%BB%83+h%C3%A3y+so+s%C3%A1nh+a+v%C3%A0+b+++a=+2020+.+2020b=+2018+.+2022+++++++++

a>b

a) \(A=2019.2021=\left(2020-1\right).\left(2020+1\right)=2020^2-1\)

\(B=2020.2020=2020^2\)

\(\Rightarrow2020^2-1< 2020^2\)\(\Rightarrow A< B\)

b) \(C=35.53-18=\left(34+1\right).53-18=34.53+53-18=34.53+34\)

mà \(D=35+53.34\)

\(\Rightarrow C=D\)

Đáp án là a>b!

Học tốt!

a) A= 2018 x 2022 và B= 2020 x 2020

ta có :

A=2018.2022=2020.2022-2.2022

=2020.2020+2020.2-2.2022

=2020.2020+2(2020-2022)

=2020.2020-4=B-4

=>A=B-4

hay B > A 4 đơn vị

Ta có: $a=2020.2020=\left(2018+2\right).2020=2018.2020+2020.2$

          $b=2018.2020=2018.\left(2020+2\right)=2018.2020+2018.2$

Vì $2018.2020=2018.2020$

Mà $2020.2>2018.2$

Nên $2018.2020+2020.2>2018.2020+2018.2$

Vậy $a>b$