Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

Xét B = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

=\(1-\frac{1}{2018}\)

Xét : \(\frac{2018}{2018}=1\)=) B < 1

khoan hình như sai đề

Trước tiên ta chứng minh bổ đề: Với x, y dương thì ta có:

\(\frac{1}{x^n}+\frac{1}{y^n}\ge\frac{2^{n+1}}{\left(x+y\right)^n}\)

Với n = 1 thì nó đúng.

Giả sử nó đúng đến \(n=k\)hay \(\frac{1}{x^k}+\frac{1}{y^k}\ge\frac{2^{k+1}}{\left(x+y\right)^k}\left(1\right)\)

Ta chứng minh nó đúng đến \(n=k+1\)hay \(\frac{1}{x^{k+1}}+\frac{1}{y^{k+1}}\ge\frac{2^{k+2}}{\left(x+y\right)^{k+1}}\left(2\right)\)

Từ (1) và (2) cái ta cần chứng minh trở thành:

\(\frac{1}{x^{k+1}}+\frac{1}{y^{k+1}}\ge\left(\frac{1}{x^k}+\frac{1}{y^k}\right)\frac{2}{\left(x+y\right)}\)

\(\Leftrightarrow\left(y-x\right)\left(y^{k+1}-x^{k+1}\right)\ge0\)(đúng)

Vậy ta có ĐPCM.

Áp dụng và bài toán ta được

\(2\left(\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}+\frac{1}{\left(c+a-b\right)^{2018}}\right)\ge\frac{2^{2019}}{2^{2018}.a^{2018}}+\frac{2^{2019}}{2^{2018}.b^{2018}}+\frac{2^{2019}}{2^{2018}.c^{2018}}\)

\(\Leftrightarrow\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}+\frac{1}{\left(c+a-b\right)^{2018}}\ge\frac{1}{a^{2018}}+\frac{1}{b^{2018}}+\frac{1}{c^{2018}}\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-2.\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=P-1\)

\(\Rightarrow\left(S-P\right)^{2018}=\left(P-1-P\right)^{2018}=\left(-1\right)^{2018}=1\)

Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}=\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1009}+\frac{1}{1010}\right)\)

\(=\frac{2019}{1.2018}+\frac{2019}{2.2017}+...+\frac{2019}{1009.1010}\)

\(=2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)\)

Do đó \(A=1.2.3....2018.2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)⋮2019\)  (đpcm)

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