Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 411

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\[\textbf{а)}\ \frac{1}{3} + \frac{5}{6} = \frac{2}{6} + \frac{5}{6} = \frac{7}{6} = 1\frac{1}{6}\ \]

\[\textbf{б)}\ \frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}\ \]

\[\textbf{в)}\ \frac{9}{10} - \frac{4}{5} = \frac{9}{10} - \frac{8}{10} = \frac{1}{10}\ \]

\[\textbf{г)}\ \frac{5}{6} - \frac{5}{12} = \frac{10}{12} - \frac{5}{12} = \frac{5}{12}\text{\ \ }\]

\[\textbf{д)}\ \frac{11}{18} + \frac{1}{6} = \frac{11}{18} + \frac{3}{18} = \frac{14}{18} = \frac{7}{9}\ \]

\[\textbf{е)}\ \frac{5}{12} + \frac{3}{4} = \frac{5}{12} + \frac{9}{12} = \frac{14}{12} =\]

\[= \frac{7}{6} = 1\frac{1}{6}\ \]

\[\textbf{ж)}\ \frac{4}{5} - \frac{7}{15} = \frac{12}{15} - \frac{7}{15} = \frac{5}{15} = \frac{1}{3}\ \]

\[\textbf{з)}\ \frac{11}{21} + \frac{3}{7} = \frac{11}{21} + \frac{9}{21} = \frac{18}{21} = \frac{6}{7}\text{\ \ }\]