89. Найдите значение выражения: a) $$6\frac{5}{6}-2\frac{1}{3}$$; б) $$9\frac{1}{15}-4\frac{7}{12}+3\frac{1}{20}$$; в) $$7\frac{5}{8}+3\frac{2}{3}-8\frac{3}{16}$$; г) $$9\frac{23}{25}-4\frac{7}{10}$$. д) $$20-(3\frac{1}{6}+2\frac{3}{4})$$; e) $$1\frac{5}{6}+(3-1\frac{7}{10})$$.

Найдем значения выражений:

1. a) $$6\frac{5}{6}-2\frac{1}{3} = \frac{6 \cdot 6 + 5}{6} - \frac{2 \cdot 3 + 1}{3} = \frac{36 + 5}{6} - \frac{6+1}{3} = \frac{41}{6} - \frac{7}{3} = \frac{41}{6} - \frac{7 \cdot 2}{3 \cdot 2} = \frac{41}{6} - \frac{14}{6} = \frac{41 - 14}{6} = \frac{27}{6} = \frac{9}{2} = 4\frac{1}{2}$$

2. б) $$9\frac{1}{15}-4\frac{7}{12}+3\frac{1}{20} = \frac{9 \cdot 15 + 1}{15} - \frac{4 \cdot 12 + 7}{12} + \frac{3 \cdot 20 + 1}{20} = \frac{135+1}{15} - \frac{48 + 7}{12} + \frac{60 + 1}{20} = \frac{136}{15} - \frac{55}{12} + \frac{61}{20} = \frac{136 \cdot 4}{15 \cdot 4} - \frac{55 \cdot 5}{12 \cdot 5} + \frac{61 \cdot 3}{20 \cdot 3} = \frac{544}{60} - \frac{275}{60} + \frac{183}{60} = \frac{544 - 275 + 183}{60} = \frac{452}{60} = \frac{113}{15} = 7\frac{8}{15}$$

3. в) $$7\frac{5}{8}+3\frac{2}{3}-8\frac{3}{16} = \frac{7 \cdot 8 + 5}{8} + \frac{3 \cdot 3 + 2}{3} - \frac{8 \cdot 16 + 3}{16} = \frac{56+5}{8} + \frac{9+2}{3} - \frac{128 + 3}{16} = \frac{61}{8} + \frac{11}{3} - \frac{131}{16} = \frac{61 \cdot 6}{8 \cdot 6} + \frac{11 \cdot 16}{3 \cdot 16} - \frac{131 \cdot 3}{16 \cdot 3} = \frac{366}{48} + \frac{176}{48} - \frac{393}{48} = \frac{366 + 176 - 393}{48} = \frac{149}{48} = 3\frac{5}{48}$$

4. г) $$9\frac{23}{25}-4\frac{7}{10} = \frac{9 \cdot 25 + 23}{25} - \frac{4 \cdot 10 + 7}{10} = \frac{225 + 23}{25} - \frac{40 + 7}{10} = \frac{248}{25} - \frac{47}{10} = \frac{248 \cdot 2}{25 \cdot 2} - \frac{47 \cdot 5}{10 \cdot 5} = \frac{496}{50} - \frac{235}{50} = \frac{496 - 235}{50} = \frac{261}{50} = 5\frac{11}{50}$$

5. д) $$20-(3\frac{1}{6}+2\frac{3}{4}) = 20 - (\frac{3 \cdot 6 + 1}{6} + \frac{2 \cdot 4 + 3}{4}) = 20 - (\frac{18 + 1}{6} + \frac{8+3}{4}) = 20 - (\frac{19}{6} + \frac{11}{4}) = 20 - (\frac{19 \cdot 2}{6 \cdot 2} + \frac{11 \cdot 3}{4 \cdot 3}) = 20 - (\frac{38}{12} + \frac{33}{12}) = 20 - \frac{38 + 33}{12} = 20 - \frac{71}{12} = \frac{20 \cdot 12}{1 \cdot 12} - \frac{71}{12} = \frac{240}{12} - \frac{71}{12} = \frac{240 - 71}{12} = \frac{169}{12} = 14\frac{1}{12}$$

6. е) $$1\frac{5}{6}+(3-1\frac{7}{10}) = 1\frac{5}{6} + (3 - \frac{1 \cdot 10 + 7}{10}) = 1\frac{5}{6} + (3 - \frac{10+7}{10}) = 1\frac{5}{6} + (3 - \frac{17}{10}) = \frac{1 \cdot 6 + 5}{6} + (\frac{3 \cdot 10}{1 \cdot 10} - \frac{17}{10}) = \frac{6+5}{6} + (\frac{30}{10} - \frac{17}{10}) = \frac{11}{6} + \frac{30 - 17}{10} = \frac{11}{6} + \frac{13}{10} = \frac{11 \cdot 5}{6 \cdot 5} + \frac{13 \cdot 3}{10 \cdot 3} = \frac{55}{30} + \frac{39}{30} = \frac{55+39}{30} = \frac{94}{30} = \frac{47}{15} = 3\frac{2}{15}$$

Ответ: a) $$4\frac{1}{2}$$; б) $$7\frac{8}{15}$$; в) $$3\frac{5}{48}$$; г) $$5\frac{11}{50}$$. д) $$14\frac{1}{12}$$; e) $$3\frac{2}{15}$$.