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$\small \left[\left(1,2\overline6-0,2\overline4\right)÷1,\overline3-\dfrac{8}{15}\right]÷0,7+\left[0,\overline8÷\left(2,\overline7-\dfrac{5}{3}\right)-0,6\overline2\right]÷0,1\overline9 =$

troviamo per prima cosa la frazione originaria dei numeri periodici mettendo a numeratore tutto il numero compresa la parte periodica e senza considerare la virgola sottraendo tutto ciò che sta davanti alla parte periodica mentre a denominatore metti tanti nove per quante cifre compongono la parte periodica e tanti zero per quante cifre compongono l'antiperiodo, se esiste, ossia ciò che sta fra la virgola e la parte periodica, segui il procedimento:

$\small = \left[\left(\dfrac{126-12}{90}-\dfrac{24-2}{90}\right)÷\dfrac{13-1}{9}-\dfrac{8}{15}\right]÷\dfrac{7}{10}+\left[\dfrac{8-0}{9}÷\left(\dfrac{27-2}{9}-\dfrac{5}{3}\right)-\dfrac{62-6}{90}\right]÷\dfrac{19-1}{90} =$

$\small = \left[\left(\dfrac{114}{90}-\dfrac{22}{90}\right)÷\dfrac{\cancel{12}^4}{\cancel9_3}-\dfrac{8}{15}\right]·\dfrac{10}{7}+\left[\dfrac{8}{9}÷\left(\dfrac{25}{9}-\dfrac{5}{3}\right)-\dfrac{\cancel{56}^{28}}{\cancel{90}_{45}}\right]÷\dfrac{\cancel{18}^1}{\cancel{90}_5} =$

$\small = \left[\dfrac{\cancel{92}^{46}}{\cancel{90}_{45}}÷\dfrac{4}{3}-\dfrac{8}{15}\right]·\dfrac{10}{7}+\left[\dfrac{8}{9}÷\left(\dfrac{25-15}{9}\right)-\dfrac{28}{45}\right]÷\dfrac{1}{5} =$

$\small = \left[\dfrac{\cancel{46}^{23}}{\cancel{45}_{15}}·\dfrac{\cancel3^1}{\cancel4_2}-\dfrac{8}{15}\right]·\dfrac{10}{7}+\left[\dfrac{8}{9}÷\dfrac{10}{9}-\dfrac{28}{45}\right]·5 =$

$\small = \left[\dfrac{23}{15}·\dfrac{1}{2}-\dfrac{8}{15}\right]·\dfrac{10}{7}+\left[\dfrac{\cancel8^4}{\cancel9_1}·\dfrac{\cancel9^1}{\cancel{10}_5}-\dfrac{28}{45}\right]·5 =$

$\small = \left[\dfrac{23}{30}-\dfrac{8}{15}\right]·\dfrac{10}{7}+\left[\dfrac{4}{1}·\dfrac{1}{5}-\dfrac{28}{45}\right]·5 =$

$\small = \left[\dfrac{23-16}{30}\right]·\dfrac{10}{7}+\left[\dfrac{4}{5}-\dfrac{28}{45}\right]·5 =$

$\small = \dfrac{\cancel7^1}{\cancel{30}_3}·\dfrac{\cancel{10}^1}{\cancel7_1}+\left[\dfrac{36-28}{45}\right]·5 =$

$\small = \dfrac{1}{3}·\dfrac{1}{1}+\dfrac{8}{\cancel{45}_9}·\cancel5^1 =$

$\small = \dfrac{1}{3}+\dfrac{8}{9}·1 =$

$\small = \dfrac{1}{3}+\dfrac{8}{9}=$

$\small = \dfrac{3+8}{9}=$

$\small = \dfrac{11}{9}$

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$\small \left[\left(2,\overline7-\dfrac{2}{3}\right)÷0,4\overline2-1,2\right]×5÷\left\{\left[1,\overline5+\dfrac{3}{2}-\left(\dfrac{1}{3}+0,\overline2\right)\right]÷0,\overline{45}\right\} =$

$\small = \left[\left(\dfrac{27-2}{9}-\dfrac{2}{3}\right)÷\dfrac{42-4}{90}-\dfrac{\cancel{12}^6}{\cancel{10}_5}\right]×5÷\left\{\left[\dfrac{15-1}{9}+\dfrac{3}{2}-\left(\dfrac{1}{3}+\dfrac{2-0}{9}\right)\right]÷\dfrac{45-0}{99}\right\} =$

$\small = \left[\left(\dfrac{25}{9}-\dfrac{2}{3}\right)÷\dfrac{\cancel{38}^{19}}{\cancel{90}_{45}}-\dfrac{6}{5}\right]×5÷\left\{\left[\dfrac{14}{9}+\dfrac{3}{2}-\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\right]÷\dfrac{\cancel{45}^5}{\cancel{99}_{11}}\right\} =$

$\small = \left[\left(\dfrac{25-6}{9}\right)÷\dfrac{19}{45}-\dfrac{6}{5}\right]×5÷\left\{\left[\dfrac{14}{9}+\dfrac{3}{2}-\left(\dfrac{3+2}{9}\right)\right]÷\dfrac{5}{11}\right\} =$

$\small = \left[\dfrac{\cancel{19}^1}{\cancel9_1}×\dfrac{\cancel{45}^5}{\cancel{19}_1}-\dfrac{6}{5}\right]×5÷\left\{\left[\dfrac{14}{9}+\dfrac{3}{2}-\dfrac{5}{9}\right]×\dfrac{11}{5}\right\} =$

$\small = \left[\dfrac{1}{1}×\dfrac{5}{1}-\dfrac{6}{5}\right]×5÷\left\{\left[\dfrac{28+27-10}{18}\right]×\dfrac{11}{5}\right\} =$

$\small = \left[5-\dfrac{6}{5}\right]×5÷\left\{\dfrac{\cancel{45}^5}{\cancel{18}_2}×\dfrac{11}{5}\right\} =$

$\small = \left[\dfrac{25-6}{5}\right]×5÷\left\{\dfrac{\cancel5^1}{2}×\dfrac{11}{\cancel5_1}\right\} =$

$\small = \dfrac{19}{\cancel5_1}×\cancel5^1÷\left\{\dfrac{1}{2}×\dfrac{11}{1}\right\} =$

$\small = 19÷\dfrac{11}{2} =$

$\small = \dfrac{38}{11} $

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$\small = \left(9,2\overline5-8,7\overline5\right)÷0,48+\left[\left(3,\overline8-\dfrac{5}{2}+1,\overline3\right)÷0,4\overline6-\dfrac{25}{6}\right]÷5-\dfrac{1}{8} =$

$\small = \left(\dfrac{925-92}{90}-\dfrac{875-87}{90}\right)÷\dfrac{\cancel{48}^{12}}{\cancel{100}_{25}}+\left[\left(\dfrac{38-3}{9}-\dfrac{5}{2}+\dfrac{13-1}{9}\right)÷\dfrac{46-4}{90}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \left(\dfrac{833}{90}-\dfrac{788}{90}\right)÷\dfrac{12}{25}+\left[\left(\dfrac{35}{9}-\dfrac{5}{2}+\dfrac{12}{9}\right)÷\dfrac{\cancel{42}^7}{\cancel{90}_{15}}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{\cancel{45}^1}{\cancel{90}_2}×\dfrac{25}{12}+\left[\left(\dfrac{70-45+24}{18}\right)÷\dfrac{7}{15}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{1}{2}×\dfrac{25}{12}+\left[\dfrac{\cancel{49}^7}{\cancel{18}_6}×\dfrac{\cancel{15}^5}{\cancel7_1}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\left[\dfrac{7}{6}×\dfrac{5}{1}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\left[\dfrac{35}{6}-\dfrac{25}{6}\right]×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\dfrac{\cancel{10}^5}{\cancel6_3}×\dfrac{1}{5}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\dfrac{\cancel5^1}{3}×\dfrac{1}{\cancel5_1}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\dfrac{1}{3}×\dfrac{1}{1}-\dfrac{1}{8} =$

$\small = \dfrac{25}{24}+\dfrac{1}{3}-\dfrac{1}{8} =$

$\small = \dfrac{25+8-3}{24} =$

$\small = \dfrac{\cancel{30}^5}{\cancel{24}_4} =$

$\small = \dfrac{5}{4} $

Quali es. ? Un solo post alla volta (vedi regolamento) svolgo l es.175: