Cho M=(1/1+1/2+1/3+...+1/2018)x2x3x4x..x2018)
Chứng minh M chia hết cho 2019.
(Ai thi Toán rồi thì câu này hơi bị quen à nha)
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M=[ 1+1/2018 +1/2 +1/2017 +1/3 +1/2016 +........+1/1009 +1/1010] .2.3.4...2018
M=[2019/2018 =2019/2.2017 +2019/3.2016 +....+2019/1009.1010].2.3.....2018
M.=2019.[1/2018 +1/2.2017 +.....+1/1009.1010] .2.3....2018 chia het cho 2019
suy ra M chia het cho2019
vay M chia het cho2019
\(S=1+2+2^2+2^3+...+2^{29}\)
\(S=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{27}+2^{28}+2^{29}\right)\)
\(S=7+2^3.\left(1+2+2^2\right)+...+2^{27}.\left(1+2+2^2\right)\)
\(S=7+2^3.7+...+2^{27}.7\)
\(S=7.\left(1+2^3+...+2^{27}\right)\)
Vì \(7⋮7\) nên \(7.\left(1+2^3+...+2^{27}\right)⋮7\)
Vậy \(S⋮7\)
______
\(2^{x+1}+2^x.3=320\)
\(=>2^x.2+2^x.3=320\)
\(=>2^x.\left(2+3\right)=320\)
\(=>2^x.5=320\)
\(=>2^x=320:5\)
\(=>2^x=64=2^6\)
\(=>x=6\)
\(#NqHahh\)
\(#Nulc`\)
a, Ta có: \(4\equiv1\left(mod3\right)\)
\(\Rightarrow4^{2018}\equiv1\left(mod3\right)\)
\(\Rightarrow4^{2018}-1⋮3\)
b, Ta có: \(5\equiv1\left(mod4\right)\)
\(\Rightarrow5^{2019}\equiv1\left(mod4\right)\)
\(\Rightarrow5^{2019}-1⋮4\)
c, \(4\equiv-1\left(mod5\right)\)
\(\Rightarrow4^{2019}\equiv-1\left(mod5\right)\)
\(\Rightarrow4^{2019}+1⋮5\)
d, \(5\equiv-1\left(mod6\right)\)
\(\Rightarrow5^{2017}\equiv-1\left(mod6\right)\)
\(\Rightarrow5^{2017}+1⋮6\)
1. Vì \(4\) chia \(3\) dư \(1\)
\(\Rightarrow4^{2018}\) chia \(3\) dư \(1^{2018}=1.\)
\(\Rightarrow4^{2018}-1\) chia hết cho \(3.\)
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\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^8+3^9\right)=\)
\(=4+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)=\)
\(=4\left(1+3^2+3^4+...+3^8\right)⋮4\)
M= ( 1/1+1/2+1/3+...+1/2018).(673.3).2.4.5....2018
M= (1/1+1/2+1/3+...+1/2018).2019.2.4.5...2018
vi bieu thuc tren co so 2019
=> M chia het cho 2019
nhầm đề r NGUYỄN BÙI KHÁNH NGỌC ạ