Bài 2:

\(A=1+2021+2021^2+\cdots+2021^{72}\)

=>\(2021A=2021+2021^2+\cdots+2021^{73}\)

=>2021A-A=\(2021+2021^2+\cdots+2021^{73}-1-2021-\cdots-2021^{72}\)

=>2020A=\(2021^{73}-1\)

=>2020A=B

=>B>A

Bài 1:

Gọi số cần tìm là x

x chia 30 dư 11

=>x-11⋮30

=>x-11+30⋮30

=>x+19⋮30(1)

x chia 39 dư 20

=>x-20⋮39

=>x-20+39⋮39

=>x+19⋮39(2)

x chia 42 dư 23

=>x-23⋮42

=>x-23+42⋮42

=>x+19⋮42(3)

Ta có: \(30=2\cdot3\cdot5;39=3\cdot13;42=2\cdot3\cdot7\)

=>BCNN(30;39;42)=\(2\cdot3\cdot5\cdot13\cdot7=2730\)

Từ (1),(2),(3) suy ra x+19∈BC(30;39;42)

=>x+19∈B(2730)

=>x+19∈{2730;5460;8190;10920;...}

=>x∈{2711;5441;8171;10901;...}

mà 1000<=x<=9999

nên x∈{2711;5441;8171}

2020.2022=(2021−1)(2021+1)=20212−1<\(2021^2\)

\(\Rightarrow2020.2021.2022< 2021^2.2021=2021^3\)

20212 cái này là \(2021^2\) nha!

\(B=\left(2021-1\right)\left(2021+1\right).2021=\left(2021^2-1\right).2021=2021^3-2021< A\)

B=(2021−1)(2021+1).2021=(20212−1).2021=20213−2021<A

\(A>\left(2021^2-1\right)\cdot2021=\left(2021-1\right)\left(2021+1\right)\cdot2021=B\)

\(B=2020.2021.2022=\left(2021-1\right).2021.\left(2021+1\right)=\left[\left(2021-1\right)\left(2021+1\right)\right].2021=\left(2021^2-2021+2021-1\right).2021=\left(2021^2-1\right).2021=2021^3-2021< 2021^3=A\)

vậy B<A

\(B=2020.2021.2022\\ B=\left(2021-1\right).2021.\left(2021+1\right)\\ B=\left(2021^2-1\right).2021\\ B=2021^3-2021\\ \Rightarrow A>B\)

Tham khảo:https://hoc24.vn/cau-hoi/cho-a-20213-va-b-202020212022-khong-tinh-cu-the-cac-gia-tri-cua-a-va-b-hay-so-sanh-a-va-b-ai-lam-giup-mik-voi.3007463332171

có sai ko bạn 

lúc mình làm xiong thấy sai

A<B

A<B nhé

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

\(A=2021^3=2021\cdot2021^2>2021\left(2021^2-1\right)=2021\left(2021\cdot2021-2021+2021-1\right)=2021\cdot\left(2021+1\right)\left(2021-1\right)=2021\cdot2022\cdot2020=B\)

\(B=2020.2021.2022=\left(2021-1\right).2021.\left(2021+1\right)=\left[\left(2021-1\right)\left(\left(2021+1\right)\right)\right].2011=\left(2021^2-1\right).2021=2021^3-2021< 2021^3=A\)