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Xét 3 số tự nhiên liên tiếp \(2020^{2021}-1;2020^{2021};2020^{2022}\) luôn có 1 số chia hết cho 3
Mà \(2020\equiv1\left(mod3\right)\Rightarrow2020^{2021}\equiv1\left(mod3\right)\)
Khi đó một trong 2 số \(2020^{2021}-1;2020^{2021}+1\) chia hết cho 3
=> đpcm
a) \(M=2020+2020^2+...+2020^{10}\)
\(M=\left(2020+2020^2\right)+\left(2020^3+2020^4\right)+...+\left(2020^9+2020^{10}\right)\)
\(M=2020\left(1+2020\right)+2020^3\left(1+2020\right)+...+2020^9\left(1+2020\right)\)
\(M=2021\left(2020+2020^3+...+2020^9\right)⋮2021\).
b) Bạn làm tương tự câu a).
b, \(A=2021+2021^2+...+2021^{2020}\)
\(=2021\left(1+2021\right)+...+2021^{2019}\left(1+2021\right)\)
\(=2022\left(2021+...+2021^{2019}\right)⋮2022\)
Vậy ta có đpcm
\(2^{2020}-2^{2017}\\ =2^{2017}\cdot2^3-2^{2017}\cdot1\\ =2^{2017}\left(2^3-1\right)\\ =2^{2017}\cdot7\)
Chia hết cho 7
a)Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
=> A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
=> A < 1 - 1/100
=> A < 99/100 < 1
b) \(A=\frac{n+3}{n-2}=\frac{n-2+5}{n-2}=1+\frac{5}{n-2}\)
Để A có giá trị nguyên <=> 5 chia hết cho n - 2
<=> n - 2 thuộc Ư(5) = {1; -1; 5; -5}
Lập bảng:
n - 2 | 1 | -1 | 5 | -5 |
n | 3 | 1 | 7 | -3 |
Vậy ....
Ta có: A = \(\frac{10^{2019}+1}{10^{2020}+1}\)
=> 10A = \(\frac{10^{2020}+10}{10^{2020}+1}=\frac{\left(10^{2020}+1\right)+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)
B = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10B = \(\frac{10^{2021}+10}{10^{2021}+1}=\frac{10^{2021}+1+9}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Do \(\frac{9}{10^{2020}+1}>\frac{9}{10^{2021}+1}\)=> \(1+\frac{9}{10^{2020}+1}>1+\frac{9}{10^{2021}+1}\)
=> 10A > 10B
=> A > B
Đặt \(A=2+2^2+2^3+...+2^{100}\)
=>\(2A=2^2+2^3+2^4+...+2^{101}\)
=>\(2A-A=A=2^{101}-2\)
Từ đây ta áp dụng t/chất đồng dư:
\(2=-1\left(mod3\right)\)(= thay cho dấu đồng dư nha)
=>\(2^{101}=\left(-1\right)^{101}\left(mod3\right)\)
=>\(2^{101}-2=-1-2=-3=0\left(mod3\right)\)
=>\(\left(2^{101}-2\right)⋮3\)
=>\(A⋮3\)
=>đpcm
Đặt A = \(\frac{2019^{2019}+1}{2019^{2020}+1}\)
=> \(2019A=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2018}{2019^{2020}+1}\)
Đặt B = \(\frac{2019^{2020}+1}{2019^{2021}+1}\)
=> \(2019B=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2018}{2019^{2021}+1}\)
Vì \(\frac{2018}{2019^{2020}+1}>\frac{2018}{2019^{2021}+1}\Rightarrow1+\frac{2018}{2019^{2020}+1}>1+\frac{2018}{2019^{2021}+1}\Rightarrow10A>10B\Rightarrow A>B\)
Bạn Đúc giúp người kiểu giì đấy :))) , giúp mà không giúp hết à ???
a) 2x + 2020 2021
=> 2x = 2021 - 2020
=> 2x = 1
=> 2x = 20
=> x = 0
b) Ta có :
4x + 14 ⋮ x + 2
=> 4. ( x + 2 ) + 6 ⋮ x + 2
Mà 4 . ( x + 2 ) ⋮ x + 2
=> 6 ⋮ x + 2 => x + 2 ∈ { 1 ; 2 ; 3 ;6 }
=> x ∈ { 0 ; 1 ; 4 } ( do x ∈ N )
c) ( x - 3 )2021 - ( x - 3 )5 = 0
=> ( x - 3 )5 . [ ( 2 - 3 )2016 - 1 ] = 0
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^5=0\\\left(x-3\right)^{2016}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{2016}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x-3\in=\left\{-1;1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x\in=\left\{2;4\right\}\end{cases}}\)
a) 2x = 2021 - 2020
2x = 1
\(\Rightarrow\)2x = 10
\(\Rightarrow\)x = 0
+) Ta có \(2^{20}=\left(2^{10}\right)^2=1024^2=\overline{...76}\).
Ta thấy \(\overline{...76}.\overline{...76}=\overline{...76}\).
Do đó \(2^{2020}=\left(2^{20}\right)^{101}=\overline{...76}\).
+) Ta có \(3^{20}=\left(3^{10}\right)^2=\left(59049\right)^2=\overline{...01}\).
Ta thấy \(\overline{...01}.\overline{...01}=\overline{...01}\).
Do dó \(8.3^{2021}=\left(3^{20}\right)^{101}.24=\overline{...01}.24=\overline{...24}\).
Vậy \(8.3^{2021}+2^{2020}=\overline{...76}+\overline{...24}=\overline{...00}⋮100\).