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МатематикаВопрос:
105. Найдите значение выражения: a) $$3 \frac{3}{7}+1 \frac{3}{14} : (8 \frac{3}{5}-1 \frac{3}{5})$$; б) $$2 \frac{5}{9} \cdot 2 \frac{1}{4}-6 \frac{2}{9} : \frac{8}{3}$$; в) $$(6 \frac{2}{5} : 2 \frac{11}{12}-16) \cdot 2 \frac{1}{4}$$; г) $$(\frac{1}{2})^3 \cdot (3 \frac{1}{3}-2 \frac{8}{9})^2$$; д) $$(3 \frac{2}{3}-2 \frac{1}{2}) \cdot 2 \frac{1}{7}-1 \frac{1}{2}$$; е) $$12 \cdot (\frac{11}{12}-\frac{9}{10}) : (3-1 \frac{1}{3})$$.
Ответ:
a)
- $$3 \frac{3}{7}+1 \frac{3}{14} : (8 \frac{3}{5}-1 \frac{3}{5}) = \frac{24}{7} + \frac{17}{14} : (\frac{43}{5} - \frac{8}{5}) = \frac{24}{7} + \frac{17}{14} : \frac{35}{5} = \frac{24}{7} + \frac{17}{14} : 7 = \frac{24}{7} + \frac{17}{14} \cdot \frac{1}{7} = \frac{24}{7} + \frac{17}{98} = \frac{24 \cdot 14}{7 \cdot 14} + \frac{17}{98} = \frac{336}{98} + \frac{17}{98} = \frac{353}{98} = 3 \frac{59}{98}$$
б)
- $$2 \frac{5}{9} \cdot 2 \frac{1}{4}-6 \frac{2}{9} : \frac{8}{3} = \frac{23}{9} \cdot \frac{9}{4} - \frac{56}{9} : \frac{8}{3} = \frac{23}{4} - \frac{56}{9} \cdot \frac{3}{8} = \frac{23}{4} - \frac{7}{3} = \frac{23 \cdot 3}{4 \cdot 3} - \frac{7 \cdot 4}{3 \cdot 4} = \frac{69}{12} - \frac{28}{12} = \frac{41}{12} = 3 \frac{5}{12}$$
в)
- $$(6 \frac{2}{5} : 2 \frac{11}{12}-16) \cdot 2 \frac{1}{4} = (\frac{32}{5} : \frac{35}{12}-16) \cdot \frac{9}{4} = (\frac{32}{5} \cdot \frac{12}{35}-16) \cdot \frac{9}{4} = (\frac{384}{175}-16) \cdot \frac{9}{4} = (\frac{384}{175} - \frac{16 \cdot 175}{175}) \cdot \frac{9}{4} = (\frac{384}{175} - \frac{2800}{175}) \cdot \frac{9}{4} = \frac{-2416}{175} \cdot \frac{9}{4} = - \frac{604 \cdot 9}{175} = - \frac{5436}{175} = -31 \frac{11}{175}$$
г)
- $$(\frac{1}{2})^3 \cdot (3 \frac{1}{3}-2 \frac{8}{9})^2 = \frac{1}{8} \cdot (\frac{10}{3} - \frac{26}{9})^2 = \frac{1}{8} \cdot (\frac{10 \cdot 3}{3 \cdot 3} - \frac{26}{9})^2 = \frac{1}{8} \cdot (\frac{30}{9} - \frac{26}{9})^2 = \frac{1}{8} \cdot (\frac{4}{9})^2 = \frac{1}{8} \cdot \frac{16}{81} = \frac{2}{81}$$
д)
- $$(3 \frac{2}{3}-2 \frac{1}{2}) \cdot 2 \frac{1}{7}-1 \frac{1}{2} = (\frac{11}{3} - \frac{5}{2}) \cdot \frac{15}{7} - \frac{3}{2} = (\frac{11 \cdot 2}{3 \cdot 2} - \frac{5 \cdot 3}{2 \cdot 3}) \cdot \frac{15}{7} - \frac{3}{2} = (\frac{22}{6} - \frac{15}{6}) \cdot \frac{15}{7} - \frac{3}{2} = \frac{7}{6} \cdot \frac{15}{7} - \frac{3}{2} = \frac{5}{2} - \frac{3}{2} = \frac{2}{2} = 1$$
е)
- $$12 \cdot (\frac{11}{12}-\frac{9}{10}) : (3-1 \frac{1}{3}) = 12 \cdot (\frac{11 \cdot 5}{12 \cdot 5}-\frac{9 \cdot 6}{10 \cdot 6}) : (\frac{9}{3}-\frac{4}{3}) = 12 \cdot (\frac{55}{60}-\frac{54}{60}) : \frac{5}{3} = 12 \cdot \frac{1}{60} : \frac{5}{3} = \frac{1}{5} \cdot \frac{3}{5} = \frac{3}{25}$$
