Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 445

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

\[\textbf{б)}\ \frac{1}{3} + \frac{2}{7} = \frac{7}{21} + \frac{6}{21} = \frac{13}{21}\ \]

\[\textbf{в)}\ \frac{2}{5} + \frac{1}{3} = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}\ \]

\[\textbf{г)}\ \frac{3}{7} + \frac{4}{9} = \frac{27}{63} + \frac{28}{63} = \frac{55}{63}\ \]

\[\textbf{д)}\ \frac{5}{9} - \frac{1}{6} = \frac{10}{18} - \frac{3}{18} = \frac{7}{18}\ \]

\[\textbf{е)}\ \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}\ \]

\[\textbf{ж)}\ \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\ \]

\[\textbf{з)}\ \frac{9}{5} - \frac{7}{10} = \frac{18}{10} - \frac{7}{10} = \frac{11}{10} =\]

\[= 1\frac{1}{10}\ \]

\[\textbf{и)}\ \frac{1}{2} - \frac{3}{8} = \frac{4}{8} - \frac{3}{8} = \frac{1}{8}\ \]

\[к)\ \frac{7}{15} - \frac{3}{10} = \frac{14}{30} - \frac{9}{30} = \frac{5}{30} = \frac{1}{6}\ \]

\[л)\ \frac{3}{8} + \frac{5}{12} = \frac{9}{24} + \frac{10}{24} = \frac{19}{24}\ \]

\[м)\ \frac{5}{9} - \frac{1}{6} = \frac{10}{18} - \frac{3}{18} = \frac{7}{18}\ \]

\[н)\ \frac{5}{11} + \frac{3}{5} = \frac{25}{55} + \frac{33}{55} = \frac{58}{55} =\]

\[= 1\frac{3}{55}\ \]

\[о)\ \frac{17}{30} - \frac{3}{6} = \frac{17}{30} - \frac{15}{30} = \frac{2}{30} = \frac{1}{15}\ \]

\[п)\ \frac{17}{35} - \frac{4}{15} = \frac{51}{105} - \frac{28}{105} = \frac{23}{105}\ \]