Решение

a) $$38\frac{1}{8}+4\frac{5}{24} = 38\frac{3}{24} + 4\frac{5}{24} = (38+4) + (\frac{3}{24} + \frac{5}{24}) = 42 + \frac{8}{24} = 42 + \frac{1}{3} = 42\frac{1}{3}$$

б) $$4\frac{19}{75}+6\frac{17}{45} = 4\frac{19 \cdot 3}{75 \cdot 3} + 6\frac{17 \cdot 5}{45 \cdot 5} = 4\frac{57}{225} + 6\frac{85}{225} = (4+6) + (\frac{57}{225} + \frac{85}{225}) = 10 + \frac{142}{225} = 10\frac{142}{225}$$

в) $$47\frac{3}{14}+1\frac{8}{21} = 47\frac{3 \cdot 3}{14 \cdot 3} + 1\frac{8 \cdot 2}{21 \cdot 2} = 47\frac{9}{42} + 1\frac{16}{42} = (47+1) + (\frac{9}{42} + \frac{16}{42}) = 48 + \frac{25}{42} = 48\frac{25}{42}$$

г) $$54\frac{3}{4}+18\frac{5}{6} = 54\frac{3 \cdot 3}{4 \cdot 3} + 18\frac{5 \cdot 2}{6 \cdot 2} = 54\frac{9}{12} + 18\frac{10}{12} = (54+18) + (\frac{9}{12} + \frac{10}{12}) = 72 + \frac{19}{12} = 72 + 1\frac{7}{12} = 73\frac{7}{12}$$

д) $$28\frac{5}{9}+13\frac{3}{4} = 28\frac{5 \cdot 4}{9 \cdot 4} + 13\frac{3 \cdot 9}{4 \cdot 9} = 28\frac{20}{36} + 13\frac{27}{36} = (28+13) + (\frac{20}{36} + \frac{27}{36}) = 41 + \frac{47}{36} = 41 + 1\frac{11}{36} = 42\frac{11}{36}$$

e) $$\frac{4}{7}+2\frac{3}{5} = \frac{4 \cdot 5}{7 \cdot 5} + 2\frac{3 \cdot 7}{5 \cdot 7} = \frac{20}{35} + 2\frac{21}{35} = 2 + (\frac{20}{35} + \frac{21}{35}) = 2 + \frac{41}{35} = 2 + 1\frac{6}{35} = 3\frac{6}{35}$$

ж) $$9 + 2\frac{2}{9} = (9+2) + \frac{2}{9} = 11\frac{2}{9}$$

з) $$3\frac{11}{24}+\frac{1}{6} = 3\frac{11}{24} + \frac{1 \cdot 4}{6 \cdot 4} = 3\frac{11}{24} + \frac{4}{24} = 3 + (\frac{11}{24} + \frac{4}{24}) = 3 + \frac{15}{24} = 3 + \frac{5}{8} = 3\frac{5}{8}$$

Ответ:

• a) $$42\frac{1}{3}$$
• б) $$10\frac{142}{225}$$
• в) $$48\frac{25}{42}$$
• г) $$73\frac{7}{12}$$
• д) $$42\frac{11}{36}$$
• e) $$3\frac{6}{35}$$
• ж) $$11\frac{2}{9}$$
• з) $$3\frac{5}{8}$$