Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Решение уравнений Задание 15

\[\boxed{\text{5.\ 15}}\]

\[\textbf{а)}\ 2\frac{3}{6} + 3\frac{5}{9} - 6\frac{3}{12} =\]

\[= 2\frac{18}{36} + 3\frac{20}{36} - 6\frac{9}{36} =\]

\[= 5\frac{38}{36} - 6\frac{9}{36} =\]

\[= - \left( 6\frac{9}{36} - 6\frac{2}{36} \right) = - \frac{7}{36}\]

\[\textbf{б)}\ 4\frac{1}{4} - 3\frac{2}{3} - 2\frac{7}{8} =\]

\[= 4\frac{6}{24} - 3\frac{16}{24} - 2\frac{21}{24} =\]

\[= 4\frac{6}{24} - 5\frac{37}{24} =\]

\[= - \left( 6\frac{13}{24} - 4\frac{6}{24} \right) = - 2\frac{7}{24}\]

\[\textbf{в)}\ 4\frac{1}{6} - 9\frac{4}{9} + 3\frac{7}{12} =\]

\[= 4\frac{6}{36} - 9\frac{16}{36} + 3\frac{21}{36} =\]

\[= 7\frac{27}{36} - 8\frac{52}{36} = - 1\frac{25}{36}\]

\[\textbf{г)}\ 7\frac{1}{5} - 4\frac{2}{15} - 9\frac{2}{3} =\]

\[= 7\frac{3}{15} - 4\frac{2}{15} - 9\frac{10}{15} =\]

\[= 7\frac{3}{15} - 13\frac{12}{15} = - 6\frac{9}{15} =\]

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