§ 2. Задание 23. «Алгебра. Рабочая тетрадь. 7 класс. В 2 частях. Часть 1» АЛГЕБРА 7 ГДЗ Задание 23

    Задание 23

    Выполните действия:

      • \({\largeа)}\ \frac{1}{33}+\frac{5}{99}=\phantom{0}\definecolor{dots}{RGB}{110,177,146}\color{dots}{\large...............}\)
      • \({\largeб)}\ \frac{1}{33}+\frac{5}{44}=\phantom{0}\color{dots}{\large...............}\)
      • \({\largeв)}\ \frac{11}{60}-\frac{1}{12}=\phantom{0}\color{dots}{\large...............}\)
      • \({\largeг)}\ \frac{5}{14}-\frac{1}{21}=\phantom{0}\color{dots}{\large...............}\)
      • \({\largeд)}\ \frac{5}{36}\cdot6=\phantom{\ }\color{dots}{\large..................}\)
      • \({\largeе)}\ 8\cdot\frac{3}{40}=\phantom{\ }\color{dots}{\large..................}\)
      • \({\largeж)}\ \frac{35}{34}:5=\phantom{\ .}\color{dots}{\large.................}\)
      • \({\largeз)}\ 25:\frac{5}{11}=\phantom{0.}\color{dots}{\large................}\)

    Источник заимствования: Алгебра. Рабочая тетрадь. 7 класс. В 2 частях. Часть 1 / – Просвещение, 2018. – 12 c. ISBN 978-5-09-051661-7
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    Решение:

      • \({\largeа)}\ \frac{1^{\backslash3}}{33\phantom{^{\backslash3}}}+\frac{5^{\backslash1}}{99\phantom{^{\backslash1}}}=\frac{3+5}{99}=\frac{8}{99};\)
      • \({\largeб)}\ \frac{1^{\backslash4}}{33\phantom{^{\backslash4}}}+\frac{5^{\backslash3}}{44\phantom{^{\backslash3}}}=\frac{4+15}{132}=\frac{19}{132};\)
      • \({\largeв)}\ \frac{11^{\backslash1}}{60\phantom{^{\backslash1}}}-\frac{1^{\backslash5}}{12\phantom{^{\backslash5}}}=\frac{11-5}{60}=\frac{6}{60}=\frac{1}{10};\)
      • \({\largeг)}\ \frac{5^{\backslash3}}{14\phantom{^{\backslash3}}}-\frac{1^{\backslash2}}{21\phantom{^{\backslash2}}}=\frac{15-2}{42}=\frac{13}{42};\)
      • \({\largeд)}\ \frac{5}{36}\cdot6=\frac{5\cdot6}{36}=\frac{5}{6};\)
      • \({\largeе)}\ 8\cdot\frac{3}{40}=\frac{8\cdot3}{40}=\frac{3}{5};\)
      • \({\largeж)}\ \frac{35}{34}:5=\frac{35\cdot1}{34\cdot5}=\frac{7}{34};\)
      • \({\largeз)}\ 25:\frac{5}{11}=\frac{25\cdot11}{5}=55.\)