Выполните действия a) $$1\frac{7}{15}+2\frac{3}{15}-1\frac{8}{15}-1\frac{4}{15}$$; б) $$3\frac{2}{17}+1\frac{12}{17}+2\frac{3}{17}-3\frac{8}{17}$$; в) $$3\frac{12}{13}-1\frac{5}{13}-1\frac{9}{13}+2\frac{3}{13}$$; г) $$5\frac{3}{14}-1\frac{5}{14}+2\frac{9}{14}-4\frac{13}{14}$$. a) $$(5-3\frac{7}{16})+(4-2\frac{9}{16})$$; б) $$(2-1\frac{5}{9})+(4-1\frac{4}{9})$$; в) $$(3-1\frac{4}{7})+(5-2\frac{5}{7})$$; г) $$(2-1\frac{5}{18})+(7-2\frac{11}{18})$$. a) $$(6-3\frac{5}{12})-(3-1\frac{1}{12})$$; б) $$(7-3\frac{7}{11})-(4-2\frac{3}{11})$$; в) $$(5-2\frac{7}{19})-(4-3\frac{2}{19})$$; г) $$(8-4\frac{13}{20})-(6-5\frac{3}{20})$$.

Выполните действия

Разберем каждый пример по порядку:

a) $$1\frac{7}{15}+2\frac{3}{15}-1\frac{8}{15}-1\frac{4}{15} = (1+2-1-1) + (\frac{7}{15}+\frac{3}{15}-\frac{8}{15}-\frac{4}{15}) = 1 + \frac{7+3-8-4}{15} = 1 + \frac{-2}{15} = 1 - \frac{2}{15} = \frac{15}{15} - \frac{2}{15} = \frac{13}{15}$$

б) $$3\frac{2}{17}+1\frac{12}{17}+2\frac{3}{17}-3\frac{8}{17} = (3+1+2-3) + (\frac{2}{17}+\frac{12}{17}+\frac{3}{17}-\frac{8}{17}) = 3 + \frac{2+12+3-8}{17} = 3 + \frac{9}{17} = 3\frac{9}{17}$$

в) $$3\frac{12}{13}-1\frac{5}{13}-1\frac{9}{13}+2\frac{3}{13} = (3-1-1+2) + (\frac{12}{13}-\frac{5}{13}-\frac{9}{13}+\frac{3}{13}) = 3 + \frac{12-5-9+3}{13} = 3 + \frac{1}{13} = 3\frac{1}{13}$$

г) $$5\frac{3}{14}-1\frac{5}{14}+2\frac{9}{14}-4\frac{13}{14} = (5-1+2-4) + (\frac{3}{14}-\frac{5}{14}+\frac{9}{14}-\frac{13}{14}) = 2 + \frac{3-5+9-13}{14} = 2 + \frac{-6}{14} = 2 - \frac{3}{7} = \frac{14}{7} - \frac{3}{7} = \frac{11}{7} = 1\frac{4}{7}$$

a) $$(5-3\frac{7}{16})+(4-2\frac{9}{16}) = (5-3-\frac{7}{16})+(4-2-\frac{9}{16}) = (2-\frac{7}{16})+(2-\frac{9}{16}) = 4 - (\frac{7}{16}+\frac{9}{16}) = 4 - \frac{16}{16} = 4 - 1 = 3$$

б) $$(2-1\frac{5}{9})+(4-1\frac{4}{9}) = (2-1-\frac{5}{9}) + (4-1-\frac{4}{9}) = (1-\frac{5}{9}) + (3-\frac{4}{9}) = 4 - (\frac{5}{9}+\frac{4}{9}) = 4 - \frac{9}{9} = 4 - 1 = 3$$

в) $$(3-1\frac{4}{7})+(5-2\frac{5}{7}) = (3-1-\frac{4}{7})+(5-2-\frac{5}{7}) = (2-\frac{4}{7}) + (3-\frac{5}{7}) = 5 - (\frac{4}{7}+\frac{5}{7}) = 5 - \frac{9}{7} = 5 - 1\frac{2}{7} = 3\frac{5}{7}$$

г) $$(2-1\frac{5}{18})+(7-2\frac{11}{18}) = (2-1-\frac{5}{18}) + (7-2-\frac{11}{18}) = (1-\frac{5}{18}) + (5-\frac{11}{18}) = 6 - (\frac{5}{18}+\frac{11}{18}) = 6 - \frac{16}{18} = 6 - \frac{8}{9} = 5\frac{1}{9}$$

a) $$(6-3\frac{5}{12})-(3-1\frac{1}{12}) = (6-3-\frac{5}{12}) - (3-1-\frac{1}{12}) = (3-\frac{5}{12}) - (2-\frac{1}{12}) = 3 - \frac{5}{12} - 2 + \frac{1}{12} = 1 + \frac{1}{12} - \frac{5}{12} = 1 - \frac{4}{12} = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$

б) $$(7-3\frac{7}{11})-(4-2\frac{3}{11}) = (7-3-\frac{7}{11}) - (4-2-\frac{3}{11}) = (4-\frac{7}{11}) - (2-\frac{3}{11}) = 4 - \frac{7}{11} - 2 + \frac{3}{11} = 2 + \frac{3}{11} - \frac{7}{11} = 2 - \frac{4}{11} = 1\frac{7}{11}$$

в) $$(5-2\frac{7}{19})-(4-3\frac{2}{19}) = (5-2-\frac{7}{19}) - (4-3-\frac{2}{19}) = (3-\frac{7}{19}) - (1-\frac{2}{19}) = 3 - \frac{7}{19} - 1 + \frac{2}{19} = 2 + \frac{2}{19} - \frac{7}{19} = 2 - \frac{5}{19} = 1\frac{14}{19}$$

г) $$(8-4\frac{13}{20})-(6-5\frac{3}{20}) = (8-4-\frac{13}{20}) - (6-5-\frac{3}{20}) = (4-\frac{13}{20}) - (1-\frac{3}{20}) = 4 - \frac{13}{20} - 1 + \frac{3}{20} = 3 + \frac{3}{20} - \frac{13}{20} = 3 - \frac{10}{20} = 3 - \frac{1}{2} = 2\frac{1}{2}$$

Ответ: См. решение

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