Номер №17 — ГДЗ по алгебре за 7 класс, Макарычев
Выполните действие:
а) \(\dfrac{5}{6} + \dfrac{1}{4}\);
б) \(\dfrac{7}{8} - \dfrac{5}{6}\);
в) \(\dfrac{3}{10} - \dfrac{4}{15}\);
г) \(5 - 3\dfrac{2}{7}\);
д) \(\dfrac{4}{9} \cdot \dfrac{3}{8}\);
е) \(\dfrac{5}{8} : \dfrac{9}{10}\);
ж) \(2\dfrac{6}{7} : 1\dfrac{3}{7}\);
з) \(6\dfrac{3}{5} \cdot 10\).
Выполняем действия с обыкновенными дробями.
а) \(\dfrac{5}{6} + \dfrac{1}{4} = \dfrac{10}{12} + \dfrac{3}{12} = \dfrac{13}{12} = 1\tfrac{1}{12}\).
б) \(\dfrac{7}{8} - \dfrac{5}{6} = \dfrac{21}{24} - \dfrac{20}{24} = \dfrac{1}{24}\).
в) \(\dfrac{3}{10} - \dfrac{4}{15} = \dfrac{9}{30} - \dfrac{8}{30} = \dfrac{1}{30}\).
г) \(5 - 3\tfrac{2}{7} = 4\tfrac{7}{7} - 3\tfrac{2}{7} = 1\tfrac{5}{7}\).
д) \(\dfrac{4}{9} \cdot \dfrac{3}{8} = \dfrac{4 \cdot 3}{9 \cdot 8} = \dfrac{12}{72} = \dfrac{1}{6}\).
е) \(\dfrac{5}{8} : \dfrac{9}{10} = \dfrac{5}{8} \cdot \dfrac{10}{9} = \dfrac{50}{72} = \dfrac{25}{36}\).
ж) \(2\tfrac{6}{7} : 1\tfrac{3}{7} = \dfrac{20}{7} : \dfrac{10}{7} = \dfrac{20}{7} \cdot \dfrac{7}{10} = \dfrac{20}{10} = 2\).
з) \(6\tfrac{3}{5} \cdot 10 = \dfrac{33}{5} \cdot 10 = \dfrac{330}{5} = 66\).
Ответ: а) \(1\tfrac{1}{12}\); б) \(\dfrac{1}{24}\); в) \(\dfrac{1}{30}\); г) \(1\tfrac{5}{7}\); д) \(\dfrac{1}{6}\); е) \(\dfrac{25}{36}\); ж) \(2\); з) \(66\).
Выполните действие:
а) \(\dfrac{5}{6} + \dfrac{1}{4}\);
б) \(\dfrac{7}{8} - \dfrac{5}{6}\);
в) \(\dfrac{3}{10} - \dfrac{4}{15}\);
г) \(5 - 3\dfrac{2}{7}\);
д) \(\dfrac{4}{9} \cdot \dfrac{3}{8}\);
е) \(\dfrac{5}{8} : \dfrac{9}{10}\);
ж) \(2\dfrac{6}{7} : 1\dfrac{3}{7}\);
з) \(6\dfrac{3}{5} \cdot 10\).
а) \(\dfrac{5}{6} + \dfrac{1}{4} = \dfrac{10}{12} + \dfrac{3}{12} = \dfrac{13}{12} = 1\tfrac{1}{12}\).
б) \(\dfrac{7}{8} - \dfrac{5}{6} = \dfrac{21}{24} - \dfrac{20}{24} = \dfrac{1}{24}\).
в) \(\dfrac{3}{10} - \dfrac{4}{15} = \dfrac{9}{30} - \dfrac{8}{30} = \dfrac{1}{30}\).
г) \(5 - 3\tfrac{2}{7} = 1\tfrac{5}{7}\).
д) \(\dfrac{4}{9} \cdot \dfrac{3}{8} = \dfrac{12}{72} = \dfrac{1}{6}\).
е) \(\dfrac{5}{8} : \dfrac{9}{10} = \dfrac{5}{8} \cdot \dfrac{10}{9} = \dfrac{25}{36}\).
ж) \(2\tfrac{6}{7} : 1\tfrac{3}{7} = \dfrac{20}{7} : \dfrac{10}{7} = 2\).
з) \(6\tfrac{3}{5} \cdot 10 = \dfrac{33}{5} \cdot 10 = 66\).
Ответ: а) \(1\tfrac{1}{12}\); б) \(\dfrac{1}{24}\); в) \(\dfrac{1}{30}\); г) \(1\tfrac{5}{7}\); д) \(\dfrac{1}{6}\); е) \(\dfrac{25}{36}\); ж) \(2\); з) \(66\).
