Решебник по математике 5 класс Виленкин ФГОС Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 491

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\[\textbf{а)}\ \frac{61}{64} - \left( \frac{7}{12} - \frac{5}{14} \right) \cdot \left( \frac{13}{16} + \frac{1}{2} \right) =\]

\[= \frac{61}{64} - \left( \frac{49}{84} - \frac{30}{84} \right) \cdot \left( \frac{13}{16} + \frac{8}{16} \right) =\]

\[= \frac{61}{64} - \frac{19}{84} \cdot \frac{21}{16} = \frac{61}{64} - \frac{19 \cdot 1}{4 \cdot 16} =\]

\[= \frac{61}{64} - \frac{19}{64} = \frac{42}{64} = \frac{21}{32}\]

\[\textbf{б)}\ \left( 1 - \frac{11}{17} \right) \cdot \left( \frac{3}{4} - \frac{5}{12} + \frac{11}{18} \right) =\]

\[= \left( \frac{17}{17} - \frac{11}{17} \right)\left( \frac{27}{36} - \frac{15}{36} + \frac{22}{36} \right) =\]

\[= \frac{6}{17} \cdot \frac{34}{36} = \frac{2}{6} = \frac{1}{3}\]

\[\textbf{в)}\ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} =\]

\[= \frac{60}{60} - \frac{30}{60} + \frac{20}{60} - \frac{15}{60} + \frac{12}{60} - \frac{10}{60} =\]

\[= \frac{37}{60}\]

\[= \frac{4}{8} + \frac{6}{12} + \frac{8}{16} + \frac{10}{20} =\]

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