Решебник по математике 5 класс Виленкин ФГОС Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 471

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\[\textbf{а)}\ \left( \frac{4}{9} + \frac{2}{9} \right) \cdot \frac{9}{13} = \frac{6}{9} \cdot \frac{9}{13} = \frac{6}{13}\]

\[\textbf{б)}\ \frac{2}{3} \cdot \left( \frac{9}{8} - \frac{3}{4} \right) = \frac{2}{3} \cdot \left( \frac{9}{8} - \frac{6}{8} \right) =\]

\[= \frac{2}{3} \cdot \frac{3}{8} = \frac{1}{4}\]

\[\textbf{в)}\ \left( \frac{9}{11} - \frac{4}{11} \right) \cdot \frac{11}{5} = \frac{5}{11} \cdot \frac{11}{5} = 1\]

\[\textbf{г)}\ 2 \cdot \frac{1}{8} + \frac{7}{12} \cdot \frac{3}{7} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\]

\[\textbf{д)}\ \left( \frac{2}{3} \right)^{2} + \frac{13}{21} \cdot \frac{7}{26} - \frac{5}{18} =\]

\[= \frac{7}{9} + \frac{1}{6} - \frac{5}{18} = \frac{14}{18} + \frac{3}{18} - \frac{5}{18} =\]

\[= \frac{12}{18} = \frac{2}{3}\]

\[\textbf{е)}\ \left( \frac{3}{7} - \frac{1}{7} \right)^{2} \cdot \frac{49}{16} + \left( \frac{1}{2} \right)^{3} =\]

\[= \left( \frac{2}{7} \right)^{2} \cdot \frac{49}{16} + \frac{1}{8} = \frac{4}{49} \cdot \frac{49}{16} + \frac{1}{8} =\]

\[= \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}\]