a) $A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}$
$A=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}$
$A=1-\frac{1}{2!}+\frac{1}{1!}-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}$
$A=1+1-\frac{1}{100!}$
$A=2-\frac{1}{100!}<2$
$B=\frac{2019.4036+4018}{2019.2019-10}$
$B=\frac{2019.4036+4018+10}{2019.2019-10+10}$
$B=\frac{2019.4036+4028}{2019.2019}$
$B>\frac{2019.4036+4038}{2019.2019}$
$B>\frac{2019.4036+2019.2}{2019.2019}$
$B>\frac{2019.(4036+2)}{2019.2019}$
$B>\frac{2019.4038}{2019.2019}=\frac{4038}{2019}=2$
Ta có: $A<2; 2<B$
$⇒A<B$
