Ta để ý rằng
$0,19981998... = 10 . 0,019981998... = 100. 0,0019981998...$
Vậy ta có
$B = \dfrac{2}{100.0,0019981998...} + \dfrac{2}{10.0,0019981998...} + \dfrac{2}{0,0019981998...}$
$= \dfrac{2}{0,0019981998} (\dfrac{1}{100} + \dfrac{1}{10} + 1)$
$= \dfrac{2}{0,0019981998} . \dfrac{111}{100}$
$= \dfrac{111}{50 . 0,0019981998...}$
$= \dfrac{111}{5.0,019981998...}$
$= \dfrac{111}{5.\dfrac{1}{10}. 0,19981998...}$
$= \dfrac{222}{0,19981998...}$
Ta đặt $A = 0,19981998...$. Khi đó
$10000A = 1998,19981998...$
Vậy
$10000A - A = 9999A = 1998,19981998... - 0,19981998.... = 1998$
Do đó $A = \dfrac{1998}{999} = \dfrac{222}{1111}$
Vậy
$B = \dfrac{222}{\dfrac{222}{1111}} = 1111$
