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

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=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 \(a>b\)

ta có : 

A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)

B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)

\(\Leftrightarrow\) B < A

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 \(D=35+53.34\)

\(\Rightarrow C=D\)

ta có A =2020 x 20192019

=> A = 2020 x 2019 x10001

=> A = (2020 x 10001) x 2019

=> A = 20202020 x2019 = B

vậy A = B
 

Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)

\(\dfrac{2019}{2020}=1-\dfrac{1}{2020}>1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

\(\dfrac{2019}{2020}>\dfrac{2018}{2019}\)