Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 415

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\[\textbf{а)}\ \frac{1}{2} + \frac{3}{4} + \frac{5}{8} = \frac{4}{8} + \frac{6}{8} + \frac{5}{8} =\]

\[= \frac{10}{8} + \frac{5}{8} = \frac{15}{8} = 1\frac{7}{8}\ \]

\[\textbf{б)}\ \frac{23}{24} - \frac{5}{12} - \frac{1}{6} =\]

\[= \frac{23}{24} - \frac{10}{24} - \frac{4}{24} = \frac{13}{24} - \frac{4}{24} =\]

\[= \frac{9}{24} = \frac{3}{8}\ \]

\[\textbf{в)}\ \frac{5}{8} + \frac{1}{3} + \frac{7}{12} = \frac{15}{24} + \frac{8}{24} + \frac{14}{24} =\]

\[= \frac{23}{24} + \frac{14}{24} = \frac{37}{24} = 1\frac{13}{24}\ \]

\[\textbf{г)}\ \frac{5}{6} - \frac{1}{8} + \frac{5}{12} = \frac{20}{24} - \frac{3}{24} + \frac{10}{24} =\]

\[= \frac{17}{24} + \frac{10}{24} = \frac{27}{24} = 1\frac{3}{24}\ \]

\[\textbf{д)}\ 3\frac{5}{7} + 4\frac{9}{14} - 2\frac{5}{21} =\]

\[= 3\frac{30}{42} + 4\frac{27}{42} - 2\frac{10}{42} =\]

\[= 7\frac{57}{42} - 2\frac{10}{42} = 8\frac{15}{42} - 2\frac{10}{42} =\]

\[= 6\frac{5}{42}\ \]

\[\textbf{е)}\ 2\frac{1}{3} + \frac{1}{5} - 1\frac{1}{8} =\]

\[= 2\frac{40}{120} + \frac{24}{120} - 1\frac{15}{120} =\]

\[= 2\frac{64}{120} - 1\frac{15}{120} = 1\frac{49}{120}\ \]