Найдите значение выражения, содержащего бесконечную периодическую дробь ((0,5+ 1/4 +0,1(6)+0,125)/(0,333...+0,4+ 14/15 ))+(((3,75-0,625)*48/125)/12,8*0,25)

[latex]\frac{0,5+\frac{1}{4}+0,1(6)+0,125}{0,333...+0,4+\frac{14}{15}} + \frac{(3,75-0,625)\cdot\frac{48}{125} }{12,8}\cdot0,25 = \\ = \frac{\frac{5}{10}+\frac{1}{4}+\frac{16-1}{90}+\frac{125}{1000}}{\frac{3}{9}+\frac{4}{10}+\frac{14}{15}} + \frac{(\frac{375}{100}-\frac{625}{1000})\cdot\frac{48}{125}}{\frac{128}{10}}\cdot\frac{25}{100}=[/latex]  [latex]= \frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}}{\frac{1}{3}+\frac{2}{5}+\frac{14}{15}} + \frac{(\frac{15}{4}-\frac{5}{8})\cdot\frac{48}{125}}{\frac{64}{5}}\cdot\frac{1}{4} = \frac{\frac{12+6+4+3}{24}}{\frac{5+6+14}{15}} + \frac{\frac{30-5}{8}\cdot\frac{48}{125}}{\frac{64}{5}}\cdot\frac{1}{4} = \\ = \frac{25\cdot15}{24\cdot25} + \frac{\frac{25}{8}\cdot\frac{48}{125}}{\frac{64}{5}}\cdot\frac{1}{4} = \frac{15}{24} + \frac{\frac{6}{5}}{\frac{64}{5}}\cdot\frac{1}{4} = [/latex]  [latex]= \frac{15}{24} + \frac{6\cdot5}{5\cdot64}\cdot\frac{1}{4} = \frac{15}{24} + \frac{3}{32}\cdot\frac{1}{4} = \frac{15}{24} + \frac{3}{128} = \frac{240+3}{384} = \frac{243}{384} = \frac{81}{128}[/latex]