ГДЗ по математике 5 класс Ерина Виленкин рабочая тетрадь Часть 1, 2 Часть 2 | Страница 31

\[Стр.\ 31.\]

\[\boxed{\mathbf{4.}}\]

\[\textbf{а)}\ 3\frac{3}{5} + \frac{1}{5} = 3 + \left( \frac{3}{5} + \frac{1}{5} \right) = 3\frac{4}{5}\]

\[\textbf{б)}\ 4\frac{3}{7} + 2\frac{4}{7} = 6\frac{7}{7} = 6 + 1 = 7\]

\[\textbf{в)}\ 3\frac{5}{6} + \frac{1}{6} = 3\frac{6}{6} = 3 + 1 = 4\]

\[\textbf{г)}\ 15\frac{3}{8} + 4\frac{2}{8} = 19\frac{5}{9}\]

\[\textbf{д)}\ \frac{7}{11} + 3\frac{2}{11} = 3 + \left( \frac{7}{11} + \frac{2}{11} \right) =\]

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

\[\textbf{е)}\ 9\frac{3}{10} + 4\frac{1}{10} = 13\frac{4}{10}\]

\[\textbf{ж)}\ 17\frac{5}{12} + 1\frac{7}{12} = 18\frac{12}{12} =\]

\[= 18 + 1 = 19\]

\[\textbf{з)}\ \frac{1}{9} + 9\frac{8}{9} = 9\frac{9}{9} = 9 + 1 = 10\]

\[\boxed{\mathbf{5.}}\]

\[\textbf{а)}\ 5\frac{3}{7} - 2\frac{2}{7} = 3 + \left( \frac{3}{7} - \frac{2}{7} \right) = 3\frac{1}{7}\]

\[\textbf{б)}\ 15\frac{3}{100} - 3 =\]

\[= (15 - 3) + \frac{3}{100} = 12\frac{3}{100}\]

\[\textbf{в)}\ 14\frac{3}{11} - \frac{3}{11} =\]

\[= 14 + \left( \frac{3}{11} - \frac{3}{11} \right) = 14\]

\[\textbf{г)}\ 7\frac{2}{13} - 7 = (7 - 7) + \frac{2}{13} = \frac{2}{13}\]

\[\textbf{д)}\ 11\frac{1}{2} - 11\frac{1}{2} = 0\]

\[\textbf{е)}\ 18\frac{4}{9} - 9\frac{3}{9} = 9 + \left( \frac{4}{9} - \frac{3}{9} \right) =\]

\[= 9\frac{1}{9}\]

\[\textbf{ж)}\ 10\frac{3}{7} - \frac{1}{7} = 10 + \left( \frac{3}{7} - \frac{1}{7} \right) =\]

\[= 10\frac{2}{7}\]

\[\textbf{з)}\ 16\frac{3}{10} - 16\frac{1}{10} = \frac{3}{10} - \frac{1}{10} =\]

\[= \frac{2}{10}\]

\[\boxed{\mathbf{6.}}\]

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

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

\[\textbf{в)}\ 5 - \frac{3}{4} = 4\frac{4}{4} - \frac{3}{4} = 4\frac{1}{4}\]

\[\textbf{г)}\ 10 - \frac{2}{9} = 9\frac{9}{9} - \frac{2}{9} = 9\frac{7}{9}\]

\[\textbf{д)}\ 11 - \frac{11}{100} = 10\frac{100}{100} - \frac{11}{100} =\]

\[= 10\frac{89}{100}\]

\[\textbf{е)}\ 5 - \frac{2}{7} = 4\frac{7}{7} - \frac{2}{7} = 4\frac{5}{7}\]

\[\textbf{ж)}\ 13 - \frac{7}{10} = 12\frac{10}{10} - \frac{7}{10} =\]

\[= 12\frac{3}{10}\]

\[\textbf{з)}\ 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}\]