| № | Пример | Решение |
|---|---|---|
| 1 | \[ \frac{3}{17} + \frac{6}{17} - \frac{9}{17} + \frac{3}{11} + \frac{5}{11} + \frac{3}{11} = \] | \[ \frac{3+6-9}{17} + \frac{3+5+3}{11} = \frac{0}{17} + \frac{11}{11} = 0 + 1 = 1 \] |
| 2 | \[ \frac{5}{12} + \frac{5}{12} + \frac{10}{12} - \frac{5}{40} + \frac{9}{40} - \frac{20}{40} = \] | \[ \frac{5+5+10}{12} + \frac{9-5-20}{40} = \frac{20}{12} + \frac{-16}{40} = \frac{5}{3} - \frac{2}{5} = \frac{5 \cdot 5 - 2 \cdot 3}{15} = \frac{25-6}{15} = \frac{19}{15} = 1 \frac{4}{15} \] |
| 3 | \[ 1 \frac{4}{5} + \frac{5}{5} + \frac{3}{8} - \frac{8}{8} = \] | \[ \frac{9}{5} + 1 + \frac{3}{8} - 1 = \frac{9}{5} + \frac{3}{8} = \frac{9 \cdot 8 + 3 \cdot 5}{40} = \frac{72 + 15}{40} = \frac{87}{40} = 2 \frac{7}{40} \] |
| 4 | \[ \frac{12}{17} - \frac{5}{17} - 2 \frac{3}{25} = \] | \[ \frac{12-5}{17} - \frac{53}{25} = \frac{7}{17} - \frac{53}{25} = \frac{7 \cdot 25 - 53 \cdot 17}{17 \cdot 25} = \frac{175 - 901}{425} = \frac{-726}{425} = -1 \frac{301}{425} \] |
| 5 | \[ \frac{11}{16} - \frac{3}{16} - \frac{5}{9} - \frac{2}{9} - \frac{3}{9} = \] | \[ \frac{11-3}{16} - \frac{5+2+3}{9} = \frac{8}{16} - \frac{10}{9} = \frac{1}{2} - \frac{10}{9} = \frac{9 - 20}{18} = \frac{-11}{18} \] |
| 6 | \[ m + \frac{2}{k} - 1 \frac{14}{k} + \frac{a}{3} + \frac{b}{3} = \] | \[ m + \frac{2}{k} - \frac{14}{k} + \frac{a+b}{3} = m - \frac{12}{k} + \frac{a+b}{3} \] |
| 7 | \[ 4 - \frac{3}{n} - \frac{5}{m} - 2 \frac{5}{m} = \] | \[ 4 - \frac{3}{n} - \frac{5}{m} - \frac{25}{m} = 4 - \frac{3}{n} - \frac{30}{m} \] |
| 8 | \[ 1 \frac{2}{3} + \frac{5}{5} + \frac{6}{13} + \frac{1}{7} + 1 \frac{4}{7} + \frac{2}{11} = \] | \[ \frac{5}{3} + 1 + \frac{6}{13} + \frac{1}{7} + \frac{11}{7} + \frac{2}{11} = \frac{5}{3} + 1 + \frac{6}{13} + \frac{12}{7} + \frac{2}{11} = \frac{5}{3} + 1 + \frac{6 \cdot 7 \cdot 11 + 12 \cdot 13 \cdot 11 + 2 \cdot 13 \cdot 7}{13 \cdot 7 \cdot 11} = \frac{5}{3} + 1 + \frac{462 + 1716 + 182}{1001} = \frac{5}{3} + 1 + \frac{2360}{1001} = \frac{5}{3} + 1 + 2 \frac{358}{1001} = \frac{5}{3} + 3 \frac{358}{1001} = 1 \frac{2}{3} + 3 \frac{358}{1001} = 4 + \frac{2}{3} + \frac{358}{1001} = 4 + \frac{2 \cdot 1001 + 358 \cdot 3}{3003} = 4 + \frac{2002 + 1074}{3003} = 4 + \frac{3076}{3003} = 4 + 1 \frac{73}{3003} = 5 \frac{73}{3003} \] |
| 9 | \[ \frac{5}{12} - 1 - 5 \frac{3}{12} - 2 : \frac{15}{6} - \frac{7}{18} = \] | \[ \frac{5}{12} - 1 - \frac{63}{12} - \frac{12}{15} - \frac{7}{18} = \frac{5}{12} - 1 - \frac{63}{12} - \frac{4}{5} - \frac{7}{18} = \frac{5-63}{12} - 1 - \frac{4}{5} - \frac{7}{18} = \frac{-58}{12} - 1 - \frac{4}{5} - \frac{7}{18} = -4 \frac{5}{6} - 1 - \frac{4}{5} - \frac{7}{18} = -5 - \frac{5}{6} - \frac{4}{5} - \frac{7}{18} = -5 - \frac{5 \cdot 15 + 4 \cdot 18 + 7 \cdot 5}{90} = -5 - \frac{75 + 72 + 35}{90} = -5 - \frac{182}{90} = -5 - 2 \frac{2}{90} = -5 - 2 \frac{1}{45} = -7 \frac{1}{45} \] |
| 10 | \[ \frac{5}{9} + \frac{5}{12} + 1 + \frac{3}{4} - 2 : \frac{10}{8} = \] | \[ \frac{5}{9} + \frac{5}{12} + 1 + \frac{3}{4} - \frac{2 \cdot 8}{10} = \frac{5}{9} + \frac{5}{12} + 1 + \frac{3}{4} - \frac{16}{10} = \frac{5}{9} + \frac{5}{12} + 1 + \frac{3}{4} - \frac{8}{5} = \frac{5 \cdot 20 + 5 \cdot 15 + 180 + 3 \cdot 45 - 8 \cdot 36}{180} = \frac{100 + 75 + 180 + 135 - 288}{180} = \frac{490 - 288}{180} = \frac{202}{180} = \frac{101}{90} = 1 \frac{11}{90} \] |
| 11 | \[ \frac{9}{20} - \frac{7}{30} + \frac{11}{12} - \frac{5}{8} = \] | \[ \frac{9 \cdot 6 - 7 \cdot 4 + 11 \cdot 10 - 5 \cdot 15}{120} = \frac{54 - 28 + 110 - 75}{120} = \frac{164 - 103}{120} = \frac{61}{120} \] |
Ответ: смотри таблицу с решениями выше