[Risolto] help

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Troviamo per prima cosa le frazioni originarie dei numeri periodici:

$5,0\overline5 = \dfrac{505-50}{90} = \dfrac{455}{90} = \dfrac{91}{18}$;

$2,7\overline5 = \dfrac{275-27}{90} = \dfrac{248}{90} = \dfrac{124}{45}$;

$1,0\overline6 = \dfrac{106-10}{90} = \dfrac{96}{90} = \dfrac{16}{15}$;

$0,5\overline6 = \dfrac{56-5}{90} = \dfrac{51}{90} = \dfrac{17}{30}$;

$1,\overline9 = \dfrac{19-1}{9} = \dfrac{18}{9} = 2$;

gli altri numeri con decimali non periodici:

$4,6 = \dfrac{46}{10} = \dfrac{23}{5}$;

$0,125 = \dfrac{125}{1000} = \dfrac{1}{8}$;

$4,15 = \dfrac{415}{100} = \dfrac{83}{20}$;

$3,75 = \dfrac{375}{100} = \dfrac{15}{4}$;

quindi sostituendo con le frazioni trovate :

$\dfrac{\big[\big(\frac{91}{18}-\frac{124}{45}\big) \colon \frac{23}{5}+\big(\frac{16}{15}-\frac{17}{30}\big)×2\big]×\frac{1}{8}}{\big(\frac{83}{20}-\frac{15}{4}+2\big) \colon \big(4+\frac{4}{5}\big)}+\big(1-\frac{1}{3}\big)^0$ =

= $\dfrac{\big[\big(\frac{455-248}{90}\big)×\frac{5}{23}+\big(\frac{32-17}{30}\big)×2\big]×\frac{1}{8}}{\big(\frac{83-75+40}{20}\big) \colon\big(\frac{20+4}{5}\big)}+1$ =

= $\dfrac{\big[\frac{23}{10}×\frac{5}{23}+\frac{1}{2}×2\big]×\frac{1}{8}}{\frac{12}{5}\colon\frac{24}{5}}+1$ =

= $\dfrac{\big[\frac{115}{230}+1\big]×\frac{1}{8}}{\frac{12}{5}×\frac{5}{24}}+1$ =

= $\dfrac{\big[\frac{1}{2}+1\big]×\frac{1}{8}}{\frac{1}{1}×\frac{1}{2}}+1$ =

= $\dfrac{\big[\frac{1+2}{2}\big]×\frac{1}{8}}{\frac{1}{2}}+1$ =

= $\dfrac{\frac{3}{2}×\frac{1}{8}}{\frac{1}{2}}+1$ =

= $\dfrac{\frac{3}{16}}{\frac{1}{2}}+1$ =

= $\dfrac{3}{16}×2+1$ =

= $\dfrac{6}{16}+1$ =

= $\dfrac{3}{8}+1$ =

= $\dfrac{3+8}{8} = \dfrac{11}{8}$