Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 212

\[\boxed{\mathbf{2.}\mathbf{112}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 1\frac{4^{\backslash 4}}{15} - \frac{17^{\backslash 3}}{20} = 1\frac{16}{60} - \frac{51}{60} =\]

\[= \frac{76}{60} - \frac{51}{60} = \frac{25}{60} = \frac{5}{12}\]

\[\textbf{б)}\ 7\frac{4^{\backslash 3}}{15} - \frac{7^{\backslash 5}}{9} = 7\frac{12}{45} - \frac{35}{45} =\]

\[= 6\frac{57}{45} - \frac{35}{45} = 6\frac{22}{45}\]

\[\textbf{в)}\ 2\frac{5^{\backslash 2}}{12} - \frac{7^{\backslash 3}}{8} = 2\frac{10}{24} - \frac{21}{24} =\]

\[= 1\frac{34}{24} - \frac{21}{24} = 1\frac{13}{24}\]

\[\textbf{г)}\ 11\frac{1^{\backslash 9}}{2} - 3\frac{11}{18} =\]

\[= 11\frac{9}{18} - 3\frac{11}{18} =\]

\[= 10\frac{27}{18} - 3\frac{11}{18} = 7\frac{16}{18} = 7\frac{8}{9}\]

\[\textbf{д)}\ 9\frac{2^{\backslash 9}}{11} - 7\frac{5^{\backslash 11}}{9} = 9\frac{18}{99} - 7\frac{55}{99} =\]

\[= 8\frac{117}{99} - 7\frac{55}{99} = 1\frac{62}{99}\]

\[\textbf{е)}\ 2\frac{6^{\backslash 2}}{25} - 1\frac{7^{\backslash 5}}{10} = 2\frac{12}{50} - 1\frac{35}{50} =\]

\[= 1\frac{62}{50} - 1\frac{35}{50} = \frac{27}{50}\]

\[\textbf{ж)}\ 3\frac{1^{\backslash 2}}{2} - 2\frac{3}{4} = 3\frac{2}{4} - 2\frac{3}{4} =\]

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

\[\textbf{з)}\ 12\frac{5^{\backslash 3}}{12} - 9\frac{13^{\backslash 2}}{18} =\]

\[= 12\frac{15}{36} - 9\frac{26}{36} =\]

\[= 11\frac{51}{36} - 9\frac{26}{36} = 2\frac{25}{36}\]