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МатематикаВопрос:
Карточка 3 Найдите значение выражения: 1. 2$$\frac{1}{2}\cdot48-3\frac{2}{3}:\frac{1}{18}+5\frac{5}{12}:\frac{7}{36}$$ 2. $$(12\frac{5}{12}+1\frac{2}{3}-3\frac{5}{6}+2\frac{3}{4}):(2\frac{1}{2}\cdot\frac{2}{5}-\frac{7}{8})$$ 3. $$11\frac{2}{5}+7\frac{1}{2}\cdot(285\frac{3}{5}:14-1\frac{23}{30}+\frac{13}{50}):(24\frac{2}{5}-10\frac{23}{100})$$
Ответ:
Решение:
- $$2\frac{1}{2}\cdot48-3\frac{2}{3}:\frac{1}{18}+5\frac{5}{12}:\frac{7}{36} = \frac{5}{2}\cdot48 - \frac{11}{3}:\frac{1}{18} + \frac{65}{12}:\frac{7}{36} = 5\cdot24 - \frac{11}{3}\cdot18 +\frac{65}{12}\cdot\frac{36}{7} = 120 - 11\cdot6 + \frac{65\cdot3}{7} = 120 - 66 + \frac{195}{7} = 54 + \frac{195}{7} = \frac{54\cdot7+195}{7} = \frac{378+195}{7} = \frac{573}{7} = 81\frac{6}{7}$$
Ответ: $$81\frac{6}{7}$$
- $$(12\frac{5}{12}+1\frac{2}{3}-3\frac{5}{6}+2\frac{3}{4}):(2\frac{1}{2}\cdot\frac{2}{5}-\frac{7}{8}) = (\frac{149}{12}+\frac{5}{3}-\frac{23}{6}+\frac{11}{4}):(\frac{5}{2}\cdot\frac{2}{5}-\frac{7}{8}) = (\frac{149}{12}+\frac{5\cdot4}{3\cdot4}-\frac{23\cdot2}{6\cdot2}+\frac{11\cdot3}{4\cdot3}):(1-\frac{7}{8}) = (\frac{149+20-46+33}{12}):(\frac{8-7}{8}) = \frac{156}{12}:\frac{1}{8} = 13\cdot8 = 104$$
Ответ: 104
- $$11\frac{2}{5}+7\frac{1}{2}\cdot(285\frac{3}{5}:14-1\frac{23}{30}+\frac{13}{50}):(24\frac{2}{5}-10\frac{23}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{1428}{5}:14-\frac{53}{30}+\frac{13}{50}):(\frac{122}{5}-\frac{1023}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{1428}{5\cdot14}-\frac{53\cdot5}{30\cdot5}+\frac{13\cdot3}{50\cdot3}):(\frac{122\cdot20}{5\cdot20}-\frac{1023}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{102}{5}-\frac{265}{150}+\frac{39}{150}):(\frac{2440-1023}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{102}{5}-\frac{226}{150}):(\frac{1417}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{102}{5}-\frac{113}{75}):(\frac{1417}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{102\cdot15}{5\cdot15}-\frac{113}{75}):(\frac{1417}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{1530-113}{75}):(\frac{1417}{100}) = \frac{57}{5} + \frac{15}{2}\cdot (\frac{1417}{75}):(\frac{1417}{100}) = \frac{57}{5} + \frac{15}{2}\cdot \frac{1417}{75}\cdot\frac{100}{1417} = \frac{57}{5} + \frac{15\cdot100}{2\cdot75} = \frac{57}{5} + \frac{1500}{150} = \frac{57}{5} + 10 = \frac{57}{5} + \frac{50}{5} = \frac{107}{5} = 21\frac{2}{5}$$
Ответ: $$21\frac{2}{5}$$
