Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 453

\[\boxed{\text{5.453.}}\]

\[\textbf{а)}\ \left( \frac{3}{8} - \frac{1}{20} \right) + \frac{7}{40} =\]

\[= \left( \frac{15}{40} - \frac{2}{40} \right) + \frac{7}{40} = \frac{13}{40} + \frac{7}{40} =\]

\[= \frac{20}{40} = \frac{1}{2}\ \]

\[\textbf{б)}\ \frac{1}{6} + \left( \frac{3}{5} - \frac{1}{3} \right) =\]

\[= \frac{5}{30} + \left( \frac{18}{30} - \frac{10}{30} \right) = \frac{5}{30} + \frac{8}{30} =\]

\[= \frac{13}{30}\]

\[\textbf{в)}\ \frac{8}{9} - \left( \frac{1}{10} + \frac{2}{5} \right) =\]

\[= \frac{8}{9} - \left( \frac{1}{10} + \frac{4}{10} \right) = \frac{8}{9} - \frac{5}{10} =\]

\[= \frac{8}{9} - \frac{1}{2} = \frac{16}{18} - \frac{9}{18} = \frac{7}{18}\]

\[\textbf{г)}\ \left( \frac{5}{8} + \frac{1}{16} \right) - \frac{9}{16} =\]

\[= \left( \frac{10}{16} + \frac{1}{16} \right) - \frac{9}{16} = \frac{11}{16} - \frac{9}{16} =\]

\[= \frac{2}{16} = \frac{1}{8}\ \]