2020 x 2020 > 2019 x 2021

\(2019\times2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2=2020\times2020\)

2020x2020>2019x2021

A = 2019 x 2021

A = 2019 x (2020 + 1)

A = 2019 x 2020 + 2019

B = 2020 x (2019 + 1)
B = 2020 x 2019 + 2020

=> B > A

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 

a) 2008 x 2012 < 2009 x 2011

b) 2019 x 2021 < 2020 x 2020

             Học tốt!!!

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 \(D=35+53.34\)

\(\Rightarrow C=D\)

Tham khảo:

A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)

10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2

B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)

10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2

10A > 2 > 10B ⇒ 10A>10B ⇒ A>B

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)