Выполните действия:
а) \(3\frac{2}{15} + 1\frac{2}{5} : \frac{1}{3} - 2\frac{1}{5} = 5\frac{2}{15}\)
1) \(1\frac{2}{5} : \frac{1}{3} = \frac{7}{5} : \frac{1}{3} = \frac{7}{5} \cdot \frac{3}{1} =  \frac{7 \cdot 3}{5 \cdot 1} = \frac{21}{5} = 4\frac{1}{5}\)
2) \(3\frac{2}{15} + 4\frac{1}{5} = 3\frac{2}{15} + 4\frac{3}{15} = 3 + 4 + \frac{2}{15} + \frac{3}{15} = 7 + \frac{2 + 3}{15} = 7\frac{5}{15} = 7\frac{1}{3}\)
3) \(7\frac{1}{3} - 2\frac{1}{5} = 7\frac{5}{15} - 2\frac{3}{15} = 7 - 2 + \frac{5 - 3}{15} = 5 + \frac{2}{15} = 5\frac{2}{15}\)

б) \((1\frac{1}{2} - \frac{1}{4}) : 3\frac{3}{4} + \frac{2}{3} = 1\)
1) \(1\frac{1}{2} - \frac{1}{4} = 1\frac{2}{4} - \frac{1}{4} = 1\frac{1}{4}\)
2) \(1\frac{1}{4} : 3\frac{3}{4} = \frac{5}{4} : \frac{15}{4} = \frac{5}{4} \cdot \frac{4}{15} =  \frac{5 \cdot 4}{4 \cdot 15} = \frac{1 \cdot 1}{1 \cdot 3} = \frac{1}{3}\)
3) \(\frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1\)

в) \(4\frac{5}{6} - \frac{5}{8} - 2\frac{1}{4} \cdot \frac{1}{6} = 3\frac{5}{6}\)
1) \(2\frac{1}{4} \cdot \frac{1}{6} = \frac{9}{4} \cdot \frac{1}{6} = \frac{9 \cdot 1}{4 \cdot 6} = \frac{3 \cdot 1}{4 \cdot 2} = \frac{3}{8}\)
2) \(4\frac{5}{6} - \frac{5}{8} = 4\frac{40}{48} - \frac{30}{48} = 4\frac{10}{48} = 4\frac{5}{24}\)
3) \(4\frac{5}{24} - \frac{3}{8} = 4\frac{5}{24} - \frac{9}{24} = 3 + 1 + \frac{5}{24} - \frac{9}{24} = 3 + \frac{24}{24} + \frac{5}{24} - \frac{9}{24} = 3 + \frac{29}{24} - \frac{9}{24} = 3 + \frac{20}{24} = 3\frac{5}{6}\)

г) \((4 - 2\frac{1}{2} \cdot \frac{3}{5}) : 3\frac{1}{3} - \frac{1}{3} = \frac{5}{12}\)
1) \(2\frac{1}{2} \cdot \frac{3}{5} = \frac{5}{2} \cdot \frac{3}{5} = \frac{5 \cdot 3}{2 \cdot 5} = \frac{1 \cdot 3}{2 \cdot 1} = \frac{3}{2} = 1\frac{1}{2} = 1,5\)
2) \(4 - 1,5 = 2,5\)
3) \(2\frac{1}{2} : 3\frac{1}{3} = \frac{5}{2} : \frac{10}{3} = \frac{5}{2} \cdot \frac{3}{10} = \frac{5 \cdot 3}{2 \cdot 10} = \frac{1 \cdot 3}{2 \cdot 2} = \frac{3}{4}\)
4) \(\frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}\)

Ответ: а) \(5\frac{2}{15}\); б) \(1\); в) \(3\frac{5}{6}\); г) \(\frac{5}{12}\).

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