Найди сумму: 38 1/5 + 4 5/24

a) $38\frac{1}{5} + 4\frac{5}{24} = 38 + \frac{1}{5} + 4 + \frac{5}{24} = (38 + 4) + (\frac{1}{5} + \frac{5}{24}) = 42 + (\frac{1*24}{5*24} + \frac{5*5}{24*5}) = 42 + (\frac{24}{120} + \frac{25}{120}) = 42 + \frac{49}{120} = 42\frac{49}{120}$ б) $4\frac{19}{75} + 6\frac{17}{45} = 4 + \frac{19}{75} + 6 + \frac{17}{45} = (4 + 6) + (\frac{19}{75} + \frac{17}{45}) = 10 + (\frac{19*3}{75*3} + \frac{17*5}{45*5}) = 10 + (\frac{57}{225} + \frac{85}{225}) = 10 + \frac{142}{225} = 10\frac{142}{225}$ в) $47\frac{3}{14} + 1\frac{8}{21} = 47 + \frac{3}{14} + 1 + \frac{8}{21} = (47 + 1) + (\frac{3}{14} + \frac{8}{21}) = 48 + (\frac{3*3}{14*3} + \frac{8*2}{21*2}) = 48 + (\frac{9}{42} + \frac{16}{42}) = 48 + \frac{25}{42} = 48\frac{25}{42}$ г) $54\frac{3}{4} + 18\frac{5}{6} = 54 + \frac{3}{4} + 18 + \frac{5}{6} = (54 + 18) + (\frac{3}{4} + \frac{5}{6}) = 72 + (\frac{3*3}{4*3} + \frac{5*2}{6*2}) = 72 + (\frac{9}{12} + \frac{10}{12}) = 72 + \frac{19}{12} = 72 + 1\frac{7}{12} = 73\frac{7}{12}$ д) $28\frac{5}{9} + 13\frac{3}{4} = 28 + \frac{5}{9} + 13 + \frac{3}{4} = (28 + 13) + (\frac{5}{9} + \frac{3}{4}) = 41 + (\frac{5*4}{9*4} + \frac{3*9}{4*9}) = 41 + (\frac{20}{36} + \frac{27}{36}) = 41 + \frac{47}{36} = 41 + 1\frac{11}{36} = 42\frac{11}{36}$ е) $4\frac{4}{7} + 2\frac{3}{5} = 4 + \frac{4}{7} + 2 + \frac{3}{5} = (4 + 2) + (\frac{4}{7} + \frac{3}{5}) = 6 + (\frac{4*5}{7*5} + \frac{3*7}{5*7}) = 6 + (\frac{20}{35} + \frac{21}{35}) = 6 + \frac{41}{35} = 6 + 1\frac{6}{35} = 7\frac{6}{35}$ ж) $9\frac{5}{9} + 2\frac{1}{2} = 9 + \frac{5}{9} + 2 + \frac{1}{2} = (9 + 2) + (\frac{5}{9} + \frac{1}{2}) = 11 + (\frac{5*2}{9*2} + \frac{1*9}{2*9}) = 11 + (\frac{10}{18} + \frac{9}{18}) = 11 + \frac{19}{18} = 11 + 1\frac{1}{18} = 12\frac{1}{18}$ з) $3\frac{11}{24} + 6\frac{1}{6} = 3 + \frac{11}{24} + 6 + \frac{1}{6} = (3 + 6) + (\frac{11}{24} + \frac{1}{6}) = 9 + (\frac{11}{24} + \frac{1*4}{6*4}) = 9 + (\frac{11}{24} + \frac{4}{24}) = 9 + \frac{15}{24} = 9 + \frac{5}{8} = 9\frac{5}{8}$