a) $\frac{1}{n}$.$\frac{1}{n+1}$=$\frac{1}{n}$-$\frac{1}{n+1}$
=>$\frac{1}{n}$.$\frac{1}{n+1}$=$\frac{1}{n(n+1)}$
Ta có: $\frac{1}{n(n+1)}$=$\frac{n+1-n}{n(n+1)}$=$\frac{n+1}{n(n+1)}$-$\frac{n}{n(n+1)}$ =$\frac{1}{n}$-$\frac{1}{n+1}$
Vậy $\frac{1}{n}$.$\frac{1}{n+1}$=$\frac{1}{n}$-$\frac{1}{n+1}$
b)A= $\frac{2}{1.2}$+$\frac{2}{2.3}$+$\frac{2}{3.4}$+$\frac{2}{4.5}$+...+$\frac{2}{2018.2019}$
Ta có: $\frac{2}{1.2}$ = $\frac{2-1}{1.2}$ = $\frac{2}{1.2}$ - $\frac{1}{1.2}$ = 1- $\frac{1}{2}$
$\frac{2}{2.3}$ = $\frac{3-2}{2.3}$ = $\frac{3}{2.3}$ - $\frac{2}{2.3}$ = $\frac{1}{2}$ - $\frac{1}{3}$
=>1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+$\frac{1}{4}$-$\frac{1}{5}$+...+$\frac{1}{2018}$- $\frac{1}{2019}$
=1-$\frac{1}{2019}$
=$\frac{2018}{2019}$
Vậy A= $\frac{2018}{2019}$