Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 409

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\[\textbf{а)}\ \frac{1}{5} + \frac{1}{7} = \frac{7}{35} + \frac{5}{35} = \frac{12}{35}\ \]

\[\textbf{б)}\ \frac{1}{4} + \frac{1}{9} = \frac{9}{36} + \frac{4}{36} = \frac{13}{36}\ \]

\[\textbf{в)}\ \frac{4}{7} + \frac{4}{9} = \frac{36}{63} + \frac{28}{63} = \frac{64}{63} = 1\frac{1}{63}\ \]

\[\textbf{г)}\ \frac{1}{2} + \frac{5}{7} = \frac{7}{14} + \frac{10}{14} = \frac{17}{14} = 1\frac{3}{14}\ \]

\[\textbf{д)}\ \frac{9}{11} + 0 = \frac{9}{11}\ \]

\[\textbf{е)}\ \frac{3}{4} - \frac{3}{5} = \frac{15}{20} - \frac{12}{20} = \frac{3}{20}\ \]

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

\[\textbf{з)}\ \frac{3}{7} - \frac{2}{9} = \frac{27}{63} - \frac{14}{63} = \frac{13}{63}\ \]

\[\textbf{и)}\ \frac{4}{9} - \frac{2}{5} = \frac{20}{45} - \frac{18}{45} = \frac{2}{45}\ \]

\[к)\ \frac{11}{13} - 0 = \frac{11}{13}\ \]

\[л)\ \frac{2}{3} + \frac{3}{5} = \frac{10}{15} + \frac{9}{15} = \frac{19}{15} = 1\frac{4}{15}\ \]

\[м)\ \frac{2}{3} + \frac{3}{12} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}\text{.\ \ }\]