Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 169

\[\boxed{\text{5.169.}}\]

\[\textbf{а)}\ \frac{4}{9} + \frac{2}{9} = \frac{4 + 2}{9} = \frac{6}{9}\ \]

\[\textbf{б)}\ \frac{1}{11} + \frac{5}{11} = \frac{1 + 5}{11} = \frac{6}{11}\ \]

\[\textbf{в)}\ \frac{12}{19} + \frac{7}{19} = \frac{12 + 7}{19} = \frac{19}{19} = 1\ \]

\[\textbf{г)}\ \frac{13}{1000} + \frac{26}{1000} = \frac{13 + 26}{1000} =\]

\[= \frac{39}{1000}\ \]

\[\textbf{д)}\ \frac{5}{7} - \frac{2}{7} = \frac{5 - 2}{7} = \frac{3}{7}\ \]

\[\textbf{е)}\ \frac{4}{6} - \frac{2}{6} = \frac{4 - 2}{6} = \frac{2}{6}\ \]

\[\textbf{ж)}\ \frac{11}{16} - \frac{4}{16} = \frac{11 - 4}{16} = \frac{7}{16}\ \]

\[\textbf{з)}\ \frac{47}{100} - \frac{26}{100} = \frac{47 - 26}{100} = \frac{21}{100}\ \]