\(M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...2018\)
\(\Rightarrow M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...673.674...2018\)
Vì \(\hept{\begin{cases}M⋮3\\M⋮673\end{cases}}\) mà \(\left(3,673\right)=1\) nên \(M⋮2019\left(đpcm\right)\)
\(M=\left[\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1008}+\frac{1}{1011}\right)+\left(\frac{1}{1009}+\frac{1}{1010}\right)\right].\)\(2.3...1008.1009.1010.1011...2017.2018\)
\(=\left(\frac{2019}{2018}+\frac{2019}{2.2017}+...+\frac{2019}{1008.1011}+\frac{2019}{1009.1010}\right).2.3...1008.1009.1010.1011...2017.2018\)
\(=2019\left(\frac{1}{2018}+\frac{1}{2.2017}+...+\frac{1}{1008.1011}+\frac{1}{1009.1010}\right).2...1008.1009.1010.1011...2017.2018\)
\(=2019.\left(2...2017+3...2016.2018+...+2.3...1007.1009.1011...2018+2.3....1008.1011...2018\right)\)
Chia hết cho 2019
❖︵Ňɠυүễη Çɦâυ Ƭυấη Ƙїệт♔:Nể phục nể phục!Đã biết \(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2018}\) là số nguyên chưa ạ
