Aritmetica

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$(1,2·1,\overline6-0,6÷0,\overline4)÷(1,8·2,\overline7-2,\overline1)+\frac{3}{20}=$

$=\left(\dfrac{\cancel{12}^6}{\cancel{10}_5}·\dfrac{16-1}{9}-\dfrac{\cancel6^3}{\cancel{10}_5}÷\dfrac{4-0}{9}\right)÷\left(\dfrac{\cancel{18}^9}{\cancel{10}_5}·\dfrac{27-2}{9}-\dfrac{21-2}{9}\right)+\dfrac{3}{20}=$

$=\left(\dfrac{6}{5}·\dfrac{\cancel{15}^5}{\cancel9_3}-\dfrac{3}{5}÷\dfrac{4}{9}\right)÷\left(\dfrac{\cancel9^1}{\cancel5_1}·\dfrac{\cancel{25}^5}{\cancel9_1}-\dfrac{19}{9}\right)+\dfrac{3}{20}=$

$=\left(\dfrac{6}{5}·\dfrac{5}{3}-\dfrac{3}{5}·\dfrac{9}{4}\right)÷\left(\dfrac{1}{1}·\dfrac{5}{1}-\dfrac{19}{9}\right)+\dfrac{3}{20}=$

$=\left(\dfrac{\cancel6^2}{\cancel5_1}·\dfrac{\cancel5^1}{\cancel3_1}-\dfrac{27}{20}\right)÷\left(5-\dfrac{19}{9}\right)+\dfrac{3}{20}=$

$=\left(\dfrac{2}{1}·\dfrac{1}{1}-\dfrac{27}{20}\right)÷\left(\dfrac{45-19}{9}\right)+\dfrac{3}{20}=$

$=\left(2-\dfrac{27}{20}\right)÷\dfrac{26}{9}+\dfrac{3}{20}=$

$=\left(\dfrac{40-27}{20}\right)·\dfrac{9}{26}+\dfrac{3}{20}=$

$=\dfrac{\cancel{13}^1}{20}·\dfrac{9}{\cancel{26}_2}+\dfrac{3}{20}=$

$=\dfrac{1}{20}·\dfrac{9}{2}+\dfrac{3}{20}=$

$=\dfrac{9}{40}+\dfrac{3}{20}=$

$=\dfrac{9+6}{40}=$

$=\dfrac{\cancel{15}^3}{\cancel{40}_8}=$

$=\dfrac{3}{8}$