Cm
1^2019+2^2019+...+16^2019 chia hết cho 17
Những câu hỏi liên quan
Cm
1^2019+2^2019+...+16^2019 chia hết cho 17
cho A = [ 1/1 + 1/2 + ....+ 1/2019 ] * [ 1*2*3*****2019]
CM A chia hết cho 2020
Chứng minh A=2019+2019^2+2019^3+2019^4+2019^5+2019^6 chia hết cho 2.
+)Ta có:\(A=2019+2019^2+2019^3+2019^4+2019^5+2019^6\)
\(\Rightarrow A=\left(2019+2019^2\right)+\left(2019^3+2019^4\right)+\left(2019^5+2019^6\right)\)
\(\Rightarrow A=\left(2019+2019^2\right)+2019^2.\left(2019+2019^2\right)+2019^4.\left(2019+2019^2\right)\)
+)Ta lại có:20192 tận cùng là 1
=>2019+20192 tân cùng là 9+1=10
=>2019+20192\(⋮2\)
\(\Rightarrow\left(2019+2019^2\right)⋮2;2019^2.\left(2019+2019^2\right)⋮2;2019^4.\left(2019+2019^2\right)⋮2\)
\(\Rightarrow A⋮2\)
Vậy \(A⋮2\left(ĐPCM\right)\)
Chúc bn học tốt
A = 2019 + 20192 + 20193 + 20194 + 20195 + 20196
A = ( 2019 + 20192 ) + ( 20193 + 20194) + ( 20195 + 20196)
A = 1 . ( 2019 + 20192 ) + 20193 . (2019 + 20192 ) + 20195 . ( 2019 + 20192 )
A = 1 . 4 078 380 + 20193 . 4 078 380 + 20195 . 4 078 380
A = 4 078 380 . ( 1 + 20193 + 20195) \(⋮2\rightarrowĐPCM\)
# HOK TỐT #
\(A=2019+2019^2+2019^3+2019^4+2019^5+2019^6\)
<=> \(A=\left(2019+2019^2\right)+\left(2019^3+2019^4\right)+\left(2019^5+2019^6\right)\)
<=>\(A=2019.\left(1+2019\right)+2019^3.\left(1+2019\right)+2019^5\left(1+2019\right)\)
<=>\(A=2019.2020+2019^3.2020+2019^5.2020\)
<=>\(A=2020.\left(2019+2019^3+2019^5\right)\)
<=>\(A=2.1010\left(2019+2019^3+2019^5\right)⋮2\)=> \(A⋮2\)
Vậy .....
chứng minh rằng nếu n là số nguyên dương thì:
2(1^2019+2^2019+3^2019+...+n^2019) chia hết cho n(n+1)
Xin chào bạn ! Mình là youtuber PUBG Takaz đây !
H=1/2019+2/2018+3/2017+...+2018/2+2019/1 chứng minh H+2019 chia hết 2020. Giups mik nha đúng mik tick cho :))))
16 tháng 1 2020 lúc 16:47
CM chia hết:
52019 - 1 chia hết cho 6.
Cho a,b € N biết 7×a^2-1 chia hết cho 7×ab-1. Tính M=a^2019-b^2019
CMR 20192019 - 1 chia hết cho 2018
\(2019^{2019}-1=2019^{2019}-1^{2019}⋮2019-1=2018⋮2018\)
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Cho x,y,z là các số tự nhiên thỏa mãn điều kiện \(\frac{x-1}{2}=\frac{y+5}{5}=\frac{z-2}{3}\) và \(3x+2y-5z+16=0\). Chứng minh rằng: \(P=x^{2019}+y^{2019}+z^{2019}\) chia hết cho 4.
\(\frac{3x-3}{6}=\frac{2y+10}{10}=\frac{5z-10}{15}=\frac{3x+2y-5z+17}{1}=\frac{3x+2y-5z+16+1}{1}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x-1}{2}=1\\\frac{y+5}{5}=1\\\frac{z-2}{3}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\\z=5\end{matrix}\right.\)
\(\Rightarrow P=3^{2019}+5^{2019}\)
Ta có \(3\equiv-1\left(mod4\right)\Rightarrow3^{2019}\equiv-1\left(mod4\right)\)
\(5\equiv1\left(mod4\right)\Rightarrow5^{2019}\equiv1\left(mod4\right)\)
\(\Rightarrow P\equiv\left(-1+1\right)\left(mod4\right)\Rightarrow P\equiv0\left(mod4\right)\Rightarrow P⋮4\)
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