Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Решение уравнений Задание 29

\[\boxed{\text{5.\ 29}}\]

\[\textbf{а)}\ 4\frac{2}{15} - 3\frac{3}{10} - 2\frac{1}{6} =\]

\[= 4\frac{4}{30} - 3\frac{9}{30} - 2\frac{5}{30} =\]

\[= 4\frac{4}{30} - 5\frac{14}{30} = - 1\frac{10}{30} = - 1\frac{1}{3}\]

\[\textbf{б)}\ 7\frac{5}{21} - 14\frac{1}{7} + 6\frac{1}{14} =\]

\[= 7\frac{10}{42} - 14\frac{6}{42} + 6\frac{3}{42} =\]

\[= 10\frac{13}{42} - 13\frac{48}{42} = - 3\frac{35}{42} =\]

\[= - 3\frac{5}{6}\]

\[\textbf{в)}\ 24\frac{2}{35} - 18\frac{5}{14} - 5\frac{3}{10} =\]

\[= 24\frac{4}{70} - 18\frac{25}{70} - 5\frac{21}{70} =\]

\[= 23\frac{74}{70} - 23\frac{46}{70} = \frac{28}{70} = \frac{2}{5}\]

\[\textbf{г)}\ 1\frac{2}{9} + 2\frac{5}{6} - 35\frac{1}{5} =\]

\[= 1\frac{20}{90} - 2\frac{75}{90} - 35\frac{18}{90} =\]

\[= 1\frac{20}{90} - 37\frac{93}{90} = - 36\frac{73}{90}\]