Giải:
Ta có:A=1.2+2.3+3.4+...+2017.2018
         3A=1.2.3 2.3.3+...+2017.2018.3
             =1.2.(3-0)+2.3.(4-1)+...+2017.2018.(2019-2016)
             =1.2.3+2.3.4+...+2017.2018.2019-1.2.0-2.3.1-...-2017.2018.1016
             =2017.2018.2019-1.2.0
             =2017.2018.2019
           =>A=2017.2018.2019/3=2018.(2017.2019)/3
            Và B=20183
/3=2018.2018.2018/3=2018.(2018.2018)/3
 Lại có: 2017.2019=2017.(2018+1)=2017.2018+2017
            2018.2018=(2017+1).2018=2017.2018+2018
Mà 2017.2018+2017<2017.2018+2018 =>2017.2019<2018.2018
    =>2018.(2017.2019)<2018.(2018.2018)
   =>A=2018.(2017.2019)/3<2018.(2018.2018)/3=B
   =>A<B

CÂU NÀY LÀ TÍNH NHANH NHÉ !

Đặt \(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)

 Biến đổi mẫu 

\(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)

\(=\left(2017+1\right)+\left(\frac{2016}{2}+1\right)+...+\left(\frac{1}{2017}+1\right)-2017\)

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

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

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