$0,5-(0,5·0,\overline{09}+0,\overline{18})=$
$= \dfrac{\cancel5^1}{\cancel{10}_2}-\left(\dfrac{\cancel5^1}{\cancel{10}_2}·\dfrac{9-0}{99}+\dfrac{18-0}{99}\right)=$
$= \dfrac{1}{2}-\left(\dfrac{1}{2}·\dfrac{\cancel9^1}{\cancel{99}_{11}}+\dfrac{\cancel{18}^2}{\cancel{99}_{11}}\right)=$
$= \dfrac{1}{2}-\left(\dfrac{1}{2}·\dfrac{1}{11}+\dfrac{2}{11}\right)=$
$= \dfrac{1}{2}-\left(\dfrac{1}{22}+\dfrac{2}{11}\right)=$
$= \dfrac{1}{2}-\left(\dfrac{1+4}{22}\right)=$
$= \dfrac{1}{2}-\dfrac{5}{22}=$
$= \dfrac{11-5}{22}=$
$= \dfrac{\cancel6^3}{\cancel{22}_{11}}=$
$=\dfrac{3}{11}$
Aiuto non va bene come titolo! Stai affogando? Vedi regolamento!
0,09 con 09 periodico, due cifre periodiche:
0,09 = 9 / 99 = 1/11;
0,18; (18 periodico);
0,18 = 18/99 = 2/11
0,5 = 5/10 = 1/2;
1/2 - ( 1/2 * 1/11 + 2/11) =
= 1/2 - (1/22 + 2/11) =
= 1/2 - ( 1/22 + 4/22) =
= 1/2 - 5/22 = 11/22 - 5/22 =
= 6/22 = 3/11.
@elenaalina ciao.