Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 209

\[\boxed{\mathbf{2.}\mathbf{209}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 4\frac{2^{\backslash 2}}{9} + 3\frac{5}{18} = 4\frac{4}{18} + 3\frac{5}{18} =\]

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

\[\textbf{б)}\ 3\frac{5^{\backslash 3}}{12} + 4\frac{7^{\backslash 4}}{9} = 3\frac{15}{36} + 4\frac{28}{36} =\]

\[= 7\frac{43}{36} = 7 + 1\frac{7}{36} = 8\frac{7}{36}\]

\[\textbf{в)}\ 7\frac{11^{\backslash 2}}{12} + 1\frac{3^{\backslash 3}}{8} =\]

\[= 7\frac{22}{24} + 1\frac{9}{24} = 8\frac{31}{24} =\]

\[= 8 + 1\frac{7}{24} = 9\frac{7}{24}\]

\[\textbf{г)}\ 4\frac{3^{\backslash 11}}{8} + 2\frac{4^{\backslash 8}}{11} =\]

\[= 4\frac{33}{88} + 2\frac{32}{88} = 6\frac{65}{88}\]

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

\[\textbf{е)}\ 7\frac{3^{\backslash 3}}{7} + \frac{5}{21} = 7\frac{9}{21} + \frac{5}{21} =\]

\[= 7\frac{14}{21} = 7\frac{2}{3}\]

\[\textbf{ж)}\ 6 + 4\frac{2}{7} = 10\frac{2}{7}\]

\[\textbf{з)}\ \frac{3^{\backslash 5}}{7} + 3\frac{4^{\backslash 7}}{5} = \frac{15}{35} + 3\frac{28}{35} =\]

\[= 3\frac{43}{35} = 3 + 1\frac{8}{35} = 4\frac{8}{35}\]