Willis Turnbull
02/25/2024 · High School

Вариант 4 1.) \( \frac{9}{5} \cdot \frac{2}{3}= \) 2.) \( \frac{15}{4}: \frac{3}{7}= \) 3.) \( \frac{1}{5}-\frac{3}{4}= \) 4.) \( \frac{1}{10}+\frac{21}{50}= \) 5.) \( \left(\frac{17}{26}+\frac{11}{13}\right) \cdot \frac{17}{6}= \)

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Давайте решим каждое из заданий по порядку. 1.) \( \frac{9}{5} \cdot \frac{2}{3} = \frac{9 \cdot 2}{5 \cdot 3} = \frac{18}{15} = \frac{6}{5} \) (после сокращения на 3). 2.) \( \frac{15}{4} : \frac{3}{7} = \frac{15}{4} \cdot \frac{7}{3} = \frac{15 \cdot 7}{4 \cdot 3} = \frac{105}{12} = \frac{35}{4} \) (после сокращения на 3). 3.) \( \frac{1}{5} - \frac{3}{4} = \frac{4}{20} - \frac{15}{20} = \frac{4 - 15}{20} = \frac{-11}{20} \). 4.) \( \frac{1}{10} + \frac{21}{50} = \frac{5}{50} + \frac{21}{50} = \frac{5 + 21}{50} = \frac{26}{50} = \frac{13}{25} \) (после сокращения на 2). 5.) \( \left(\frac{17}{26} + \frac{11}{13}\right) \cdot \frac{17}{6} \). Сначала найдем сумму в скобках: \( \frac{11}{13} = \frac{22}{26} \), поэтому: \( \frac{17}{26} + \frac{22}{26} = \frac{39}{26} = \frac{3}{2} \) (после сокращения на 13). Теперь умножим: \( \frac{3}{2} \cdot \frac{17}{6} = \frac{3 \cdot 17}{2 \cdot 6} = \frac{51}{12} \). Таким образом, ответы на задания: 1.) \( \frac{6}{5} \) 2.) \( \frac{35}{4} \) 3.) \( \frac{-11}{20} \) 4.) \( \frac{13}{25} \) 5.) \( \frac{51}{12} \)

Quick Answer

1.) \( \frac{6}{5} \) 2.) \( \frac{35}{4} \) 3.) \( \frac{-11}{20} \) 4.) \( \frac{13}{25} \) 5.) \( \frac{51}{12} \)
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