Номер №160 — ГДЗ по алгебре за 7 класс, Макарычев
Выполните действия:
а) \(\left(3\dfrac{7}{30} - 1\dfrac{5}{12}\right) : 18\dfrac{1}{6}\);
б) \(\left(1\dfrac{1}{2} + 2\dfrac{2}{3}\right) : 1\dfrac{2}{3}\);
в) \(\left(\dfrac{11}{18} - 1\dfrac{7}{12}\right) \cdot \left(2\dfrac{1}{6} + \dfrac{7}{30}\right)\);
г) \(\left(3\dfrac{2}{5} - 5\right) \cdot \left(\dfrac{31}{48} + \dfrac{7}{24}\right)\).
Выполним действия в скобках, а затем умножение или деление обыкновенных дробей.
а) \[\begin{aligned} \left(3\tfrac{7}{30} - 1\tfrac{5}{12}\right) : 18\tfrac{1}{6} &= \left(3\tfrac{14}{60} - 1\tfrac{25}{60}\right) : \tfrac{109}{6} \\ &= 1\tfrac{49}{60} : \tfrac{109}{6} = \tfrac{109}{60} \cdot \tfrac{6}{109} = \tfrac{1}{10} = 0{,}1. \end{aligned}\]
б) \[\begin{aligned} \left(1\tfrac{1}{2} + 2\tfrac{2}{3}\right) : 1\tfrac{2}{3} &= \left(1\tfrac{3}{6} + 2\tfrac{4}{6}\right) : \tfrac{5}{3} \\ &= 4\tfrac{1}{6} : \tfrac{5}{3} = \tfrac{25}{6} \cdot \tfrac{3}{5} = \tfrac{5}{2} = 2{,}5. \end{aligned}\]
в) \[\begin{aligned} \left(\tfrac{11}{18} - 1\tfrac{7}{12}\right) \cdot \left(2\tfrac{1}{6} + \tfrac{7}{30}\right) &= \left(\tfrac{22}{36} - \tfrac{57}{36}\right) \cdot \left(\tfrac{65}{30} + \tfrac{7}{30}\right) \\ &= \left(-\tfrac{35}{36}\right) \cdot \tfrac{72}{30} = -\tfrac{35 \cdot 72}{36 \cdot 30} = -\tfrac{7}{3} = -2\tfrac{1}{3}. \end{aligned}\]
г) \[\begin{aligned} \left(3\tfrac{2}{5} - 5\right) \cdot \left(\tfrac{31}{48} + \tfrac{7}{24}\right) &= \left(\tfrac{17}{5} - \tfrac{25}{5}\right) \cdot \left(\tfrac{31}{48} + \tfrac{14}{48}\right) \\ &= \left(-\tfrac{8}{5}\right) \cdot \tfrac{45}{48} = -\tfrac{8 \cdot 45}{5 \cdot 48} = -\tfrac{3}{2} = -1{,}5. \end{aligned}\]
Ответ: а) \(0{,}1\); б) \(2{,}5\); в) \(-2\tfrac{1}{3}\); г) \(-1{,}5\).
Выполните действия:
а) \(\left(3\dfrac{7}{30} - 1\dfrac{5}{12}\right) : 18\dfrac{1}{6}\);
б) \(\left(1\dfrac{1}{2} + 2\dfrac{2}{3}\right) : 1\dfrac{2}{3}\);
в) \(\left(\dfrac{11}{18} - 1\dfrac{7}{12}\right) \cdot \left(2\dfrac{1}{6} + \dfrac{7}{30}\right)\);
г) \(\left(3\dfrac{2}{5} - 5\right) \cdot \left(\dfrac{31}{48} + \dfrac{7}{24}\right)\).
а) \(\left(3\tfrac{7}{30} - 1\tfrac{5}{12}\right) : 18\tfrac{1}{6} = 1\tfrac{49}{60} : \tfrac{109}{6} = \tfrac{109}{60} \cdot \tfrac{6}{109} = \tfrac{1}{10}\).
б) \(\left(1\tfrac{1}{2} + 2\tfrac{2}{3}\right) : 1\tfrac{2}{3} = \tfrac{25}{6} : \tfrac{5}{3} = \tfrac{5}{2} = 2{,}5\).
в) \(\left(\tfrac{11}{18} - 1\tfrac{7}{12}\right) \cdot \left(2\tfrac{1}{6} + \tfrac{7}{30}\right) = \left(-\tfrac{35}{36}\right) \cdot \tfrac{72}{30} = -\tfrac{7}{3} = -2\tfrac{1}{3}\).
г) \(\left(3\tfrac{2}{5} - 5\right) \cdot \left(\tfrac{31}{48} + \tfrac{7}{24}\right) = \left(-\tfrac{8}{5}\right) \cdot \tfrac{45}{48} = -\tfrac{3}{2} = -1{,}5\).
Ответ: а) \(0{,}1\); б) \(2{,}5\); в) \(-2\tfrac{1}{3}\); г) \(-1{,}5\).
