a<b nha bn

\(B=2018.2022=\left(2020-2\right)\left(2020+2\right)=2020^2-2.2020+2.2020-2.2\)

\(=A-4< A\).

a là 2020x2020 và b là 2018x2022 à bạn

nếu vậy thì b=2022x2018=(2020+2)x(2020-2)=2020x2020-4<a nha

chúc bạn học tốt

HYC-23/1/2022

b=2018.2022=(2020-2)(2020+2)=2020.2020+2.2020-2.2020-4=2020.2020-4<2020.2020

nên a>b

\(A=2020.2020=2020^2\)

\(B=\left(2020+2\right).\left(2020-2\right)=2020^2-2^2\)

Mà 2020² và 2² không âm nên

\(A>B\)

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 \(a>b\)

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

https://olm.vn/hoi-dap/question/102758.html

a 

\(B=2020^2=2020^2-1^2+1^2=\left(2020-1\right)\left(2020+1\right)+1=2019\cdot2021+1\)   

Vậy a < B 

b 

\(B=2008\cdot2012=\left(2010-2\right)\cdot\left(2010+2\right)=2010^2-2^2\)   

\(N=2009\cdot2011=\left(2010-1\right)\left(2010+1\right)=2010^2-1^2\)   

Vậy B < N 

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 $a>b$

B = 2020.2021.2022

B = (2021 - 1).(2021 + 1).2021

B = (2021.2021 + 2021 - 2021 - 1).2021

B = (2021²⁰²¹-1).2021

B = 2021³ - 2021

Vậy A > B

`a, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`

`= 3020 xx 3109 + 3015 = B`.

`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`