$A = \dfrac{10^5+4}{10^5-1}$
$ = \dfrac{(10^5-1)+5}{10^5-1}$
$ = 1+\dfrac{5}{10^5-1}$
$B = \dfrac{10^5+3}{10^5-2}$
$ = \dfrac{(10^5-2)+5}{10^5-2}$
$ = 1+\dfrac{5}{10^5-2}$
Vì $\dfrac{5}{10^5-1} < \dfrac{5}{10^5-2}$
$\to \dfrac{5}{10^5-1}+1 < \dfrac{5}{10^5-2}+1$
Hay $A<B$