6.59 Найдите разность: a) $$7 \frac{3}{5} - 1 \frac{2}{5}$$; б) $$28 \frac{6}{13} - 7 \frac{1}{13}$$; в) $$6 - 2 \frac{5}{9}$$; г) $$21 - 5 \frac{9}{9}$$; д) $$4 \frac{1}{7} - 1 \frac{4}{7}$$; e) $$12 \frac{2}{5} - 1 \frac{4}{5}$$.

Решение: a) $$7 \frac{3}{5} - 1 \frac{2}{5} = (7-1) + (\frac{3}{5} - \frac{2}{5}) = 6 + \frac{1}{5} = 6 \frac{1}{5}$$ б) $$28 \frac{6}{13} - 7 \frac{1}{13} = (28-7) + (\frac{6}{13} - \frac{1}{13}) = 21 + \frac{5}{13} = 21 \frac{5}{13}$$ в) $$6 - 2 \frac{5}{9} = (5 - 2) + (1 - \frac{5}{9}) = 3 + \frac{9-5}{9} = 3 + \frac{4}{9} = 3 \frac{4}{9}$$ г) $$21 - 5 \frac{9}{9} = 21 - 5 - \frac{9}{9} = 21 - 5 - 1 = 15$$ д) $$4 \frac{1}{7} - 1 \frac{4}{7} = 4 \frac{1}{7} - 1 - \frac{4}{7} = (4-1) + (\frac{1}{7} - \frac{4}{7}) = 3 + (\frac{1-4}{7}) = 3 - \frac{3}{7} = 2 + \frac{7-3}{7} = 2 + \frac{4}{7} = 2 \frac{4}{7}$$ e) $$12 \frac{2}{5} - 1 \frac{4}{5} = 12 \frac{2}{5} - 1 - \frac{4}{5} = (12-1) + (\frac{2}{5} - \frac{4}{5}) = 11 + (\frac{2-4}{5}) = 11 - \frac{2}{5} = 10 + \frac{5-2}{5} = 10 + \frac{3}{5} = 10 \frac{3}{5}$$