Tính giá trị biểu thức sau: B= $\frac{2}{0,19981998…}$+$\frac{2}{0,019981998…}$+$\frac{2}{0,0019981998…}$ 29/08/2021 Bởi Valentina Tính giá trị biểu thức sau: B= $\frac{2}{0,19981998…}$+$\frac{2}{0,019981998…}$+$\frac{2}{0,0019981998…}$
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 $B = \dfrac{222}{\dfrac{222}{1111}} = 1111$ Bình luận
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
$B = \dfrac{222}{\dfrac{222}{1111}} = 1111$