Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tham khảo
\(\text{+)}\)Ta có:\(5\equiv-1\left(mod3\right)\)
\(\Rightarrow5^{2022}\equiv\left(-1\right)^{2022}\left(mod3\right)\left(1\right)\)
\(\text{+)}\)Ta có:\(2\equiv-1\left(mod3\right)\)
\(\Rightarrow2^{2023}\equiv\left(-1\right)^{2023}\left(mod3\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow5^{2022}+5^{2023}\equiv0\left(mod3\right)\)
Vậy...
\(B=2021\cdot1\cdot2\cdot3\cdot...\cdot2022\cdot\left(1+\dfrac{1}{2}+...+\dfrac{1}{2022}\right)⋮2021\)
Ta có: \(\frac{2022}{2021^2+k}\le\frac{2022}{2021^2}\) (với \(k\)là số tự nhiên bất kì)
Ta có:
\(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)
\(\le\frac{2022}{2021^2}+\frac{2022}{2021^2}+...+\frac{2022}{2021^2}=\frac{2022}{2021^2}.2021=\frac{2022}{2021}\)
Ta có: \(\frac{2022}{2021^2+k}>\frac{2022}{2021^2+2021}=\frac{2022}{2021.2022}=\frac{1}{2021}\)với \(k\)tự nhiên, \(k< 2021\))
Suy ra \(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)
\(>\frac{1}{2021}+\frac{1}{2021}+...+\frac{1}{2021}=\frac{2021}{2021}=1\)
Suy ra \(1< A\le\frac{2022}{2021}\)do đó \(A\)không phải là số tự nhiên.
Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$
$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$
$\Rightarrow 5a-a=5^{2024}-1$
$\Rightarrow 4a=5^{2024}-1$
$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)
\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
A = 1 + 2022 + 20222 + 20223 + 20224 + 20225 + 20226 + 20227
A =(1 + 2022) +(20222 + 20223) + (20224 + 20225) + (20226 + 20227)
A =2023 + 20222( 1 + 2022 ) + 20224.( 1 + 2022 ) + 20226( 1 + 2022 )
A = 2023 + 20222 . 2023 + 20224 . 2023 + 20226 . 2023
A = 2023 . ( 1 + 20222 + 20224 + 20226 )
Vậy A ⋮ 2023
A = 1 + 2022 + 20222 + 20223 + 20224 + 20225 + 20226 + 20227
A =(1 + 2022) +(20222 + 20223) + (20224 + 20225) + (20226 + 20227)
A =2023 + 20222( 1 + 2022 ) + 20224.( 1 + 2022 ) + 20226( 1 + 2022 )
A = 2023 + 20222 . 2023 + 20224 . 2023 + 20226 . 2023
A = 2023 . ( 1 + 20222 + 20224 + 20226 )
Vậy A ⋮ 2023