3. Выполните действия: a) \frac{18}{27} + \frac{15}{27} - \frac{11}{27}; б) \frac{7}{8} - (\frac{1}{4} + \frac{1}{6}); д) 4 \frac{4}{10} - 2 \frac{5}{41} + 6 \frac{3}{41}; з) 1 - \frac{8}{15} \cdot (6 - 2 \frac{1}{10}) \cdot \frac{4}{13}; ж) 7 \frac{5}{9} - 4 \frac{4}{9} : \frac{18}{35} \cdot \frac{9}{14}

Краткое пояснение: Выполнение арифметических действий с дробями.

Решение:

a) \(\frac{18}{27} + \frac{15}{27} - \frac{11}{27} = \frac{18+15-11}{27} = \frac{22}{27}\)

б) \(\frac{7}{8} - (\frac{1}{4} + \frac{1}{6}) = \frac{7}{8} - (\frac{3}{12} + \frac{2}{12}) = \frac{7}{8} - \frac{5}{12} = \frac{21}{24} - \frac{10}{24} = \frac{11}{24}\)

д) \(4 \frac{4}{10} - 2 \frac{5}{41} + 6 \frac{3}{41} = 4 \frac{2}{5} + (6 \frac{3}{41} - 2 \frac{5}{41}) = 4 \frac{2}{5} + 3 \frac{39}{41} - 3 \frac{1}{205} = 4 \frac{2}{5} + 4 - \frac{2}{41} = 4 + (3 \frac{44}{41} - 2 \frac{5}{41}) = 4 \frac{2}{5} + 3 \frac{39}{41} = 4 \frac{2}{5} + 3 \frac{39}{41} = \frac{22}{5} + \frac{162}{41} = \frac{22 \cdot 41 + 162 \cdot 5}{5 \cdot 41} = \frac{902 + 810}{205} = \frac{1712}{205} = 8 \frac{72}{205}\)

з) \(1 - \frac{8}{15} \cdot (6 - 2 \frac{1}{10}) \cdot \frac{4}{13} = 1 - \frac{8}{15} \cdot (6 - \frac{21}{10}) \cdot \frac{4}{13} = 1 - \frac{8}{15} \cdot (\frac{60 - 21}{10}) \cdot \frac{4}{13} = 1 - \frac{8}{15} \cdot \frac{39}{10} \cdot \frac{4}{13} = 1 - \frac{8 \cdot 39 \cdot 4}{15 \cdot 10 \cdot 13} = 1 - \frac{2 \cdot 1 \cdot 4}{5 \cdot 5 \cdot 1} = 1 - \frac{8}{25} = \frac{25-8}{25} = \frac{17}{25}\)

ж) \(7 \frac{5}{9} - 4 \frac{4}{9} : \frac{18}{35} \cdot \frac{9}{14} = 7 \frac{5}{9} - \frac{40}{9} : \frac{18}{35} \cdot \frac{9}{14} = \frac{68}{9} - \frac{40}{9} \cdot \frac{35}{18} \cdot \frac{9}{14} = \frac{68}{9} - \frac{40 \cdot 35 \cdot 9}{9 \cdot 18 \cdot 14} = \frac{68}{9} - \frac{10 \cdot 5 \cdot 1}{1 \cdot 9 \cdot 2} = \frac{68}{9} - \frac{50}{9} = \frac{68-50}{9} = \frac{18}{9} = 2\)