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Esce così pure a me: che c'entra l'astronomia?

(5/10)^3/(36/99 + (15/99 + (43 - 4)/9 - (216 - 21)/90)·(11/27) - 21/22) - 75/1000=

=1/8/(4/11 + (5/33 + 13/3 - 13/6)·(11/27) - 21/22) - 3/40=

=1/8/(4/11 + 51/22·(11/27) - 21/22) - 3/40=

=1/8/(4/11 + 17/18 - 21/22) - 3/40=

=1/8/(35/99) - 3/40=

=99/280 - 3/40= 39/140

0,5 = 5/10 = 1/2;

0,36 (con 36 periodico) = 36/99=  4/11;

0,15 (con 15 periodico) = 15/99 = 5/33; (semplificando per 3).

4,3 con 3 periodico) = (43 - 4) / 9 = 39/9 = 13/3;

2,16 (con 6 periodico) = (216 - 21) / 90 = 195/90 = 13/6; (semplificando per 15).

0,075 = 75/1000 = 3/40; (semplificando per 25).

(1/2)^3 :[4/11 + (5/33 + 13/3 - 13/6) * 11/27 - 21/22] - 3/40 =

= 1/8 : [ 4/11 + (10/66 + 286/66 - 143/66) * 11/27 - 21/22] - 3/40 =

= 1/8 : [ 4/11 + 153/66 * 11/27 - 21/22] - 3/40 =

= 1/8 : [ 4/11 + 51/22 * 11/27 - 21/22] - 3/40 =

= 1/8 : [ 4/11 + 17/18 - 21/22] - 3/40 =

= 1/8 : [72/198 + 187/198 - 189/198 ] - 3/40 =

= 1/8 : [70/198] - 3/40 =

= 1/8 : [35/99] - 3/40 =

= 1/8 * 99/35 - 3/40 =

= 99/280 - 3/40 =

= 99/280 - 21 /280 = 

= 78/280 = 39/140.

ho corretto il mio errore. Non si distinguono i segni, scrivi meglio!

Ciao @maiscia81

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$\small \left(0,5\right)^3 : \left[0,\overline{36}+\left(0,\overline{15}+4,\overline3-2,1\overline6\right)·\dfrac{11}{27}-\dfrac{21}{22}\right]-0,075=$

$\small =\left(\dfrac{\cancel5^1}{\cancel{10}_2}\right)^3 : \left[\dfrac{36-0}{99}+\left(\dfrac{15-0}{99}+\dfrac{43-4}{9}-\dfrac{216-21}{90}\right)·\dfrac{11}{27}-\dfrac{21}{22}\right]-\dfrac{75}{1000}=$

$\small =\left(\dfrac{1}{2}\right)^3 : \left[\dfrac{\cancel{36}^4}{\cancel{99}_{11}}+\left(\dfrac{\cancel{15}^5}{\cancel{99}_{33}}+\dfrac{\cancel{39}^{13}}{\cancel9_3}-\dfrac{\cancel{195}^{13}}{\cancel{90}_6}\right)·\dfrac{11}{27}-\dfrac{21}{22}\right]-\dfrac{\cancel{75}^3}{\cancel{1000}_{40}}=$

$\small = \dfrac{1}{8} : \left[\dfrac{4}{11}+\left(\dfrac{5}{33}+\dfrac{13}{3}-\dfrac{13}{6}\right)·\dfrac{11}{27}-\dfrac{21}{22}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \left[\dfrac{4}{11}+\left(\dfrac{10+286-143}{66}\right)·\dfrac{11}{27}-\dfrac{21}{22}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \left[\dfrac{4}{11}+\dfrac{\cancel{153}^{17}}{\cancel{66}_6}·\dfrac{\cancel{11}^1}{\cancel{27}_3}-\dfrac{21}{22}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \left[\dfrac{4}{11}+\dfrac{17}{6}·\dfrac{1}{3}-\dfrac{21}{22}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \left[\dfrac{4}{11}+\dfrac{17}{18}-\dfrac{21}{22}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \left[\dfrac{504+1309-1323}{1386}\right]-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \dfrac{\cancel{490}^{35}}{\cancel{1386}_{99}}-\dfrac{3}{40}=$

$\small = \dfrac{1}{8} : \dfrac{35}{99}-\dfrac{3}{40}=$

$\small = \dfrac{1}{8}·\dfrac{99}{35}-\dfrac{3}{40}=$

$\small = \dfrac{99}{280}-\dfrac{3}{40}=$

$\small = \dfrac{99-21}{280}=$

$\small = \dfrac{\cancel{78}^{39}}{\cancel{280}_{140}}= \dfrac{39}{140}$