ГДЗ по математике 5 класс Ерина рабочая тетрадь Часть 1, 2 Часть 2 | Страница 27

Авторы:
Год:2022
Тип:рабочая тетрадь
Часть:1, 2
Серия:Учебно-методический комплект
Нужно другое издание?

Страница 27

\[29.\ Сложение\ и\ вычитание\ смешанных\ чисел\]

\[Страница\ 27.\]

\[\boxed{\mathbf{1.}}\]

\[\textbf{а)}\ 8 = 7\frac{6}{6}\]

\[\textbf{б)}\ 9 = 8\frac{4}{4}\]

\[\textbf{в)}\ 14 = 13\frac{10}{10}\]

\[\textbf{г)}\ 9 = 8\frac{5}{5}\]

\[\textbf{д)}\ 6 = 5\frac{3}{3}\]

\[\textbf{е)}\ 12 = 11\frac{7}{7}\]

\[\textbf{ж)}\ 15 = 14\frac{8}{8}\]

\[\textbf{з)}\ 24 = 23\frac{11}{11}\]

\[\textbf{и)}\ 3 = 2\frac{4}{4}\]

\[к)\ 9 = 8\frac{7}{7}\]

\[л)\ 28 = 27\frac{10}{10}\]

\[м)\ 40 = 39\frac{21}{21}\]

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

\[\textbf{а)}\ 7\frac{2}{3} = 6\frac{5}{3}\]

\[\textbf{б)}\ 15\frac{7}{9} = 14\frac{16}{9}\]

\[\textbf{в)}\ 12\frac{5}{8} = 11\frac{13}{8}\]

\[\textbf{г)}\ 1\frac{6}{10} = \frac{16}{10}\]

\[\textbf{д)}\ 11\frac{2}{7} = 10\frac{9}{7}\]

\[\textbf{е)}\ 16\frac{1}{5} = 15\frac{6}{5}\]

\[\textbf{ж)}\ 3\frac{6}{7} = 2\frac{13}{7}\]

\[\textbf{з)}\ 2\frac{8}{9} = 1\frac{17}{9}\]

\[\textbf{и)}\ 1\frac{1}{7} = \frac{8}{7}\]

\[к)\ 1\frac{3}{5} = \frac{8}{5}\]

\[л)\ 1\frac{4}{9} = \frac{13}{9}\]

\[м)\ 1\frac{9}{100} = \frac{109}{100}\]

\[Страница\ 28.\]

\[\boxed{\mathbf{3.}}\]

\[\textbf{в)}\ 13\frac{3}{7} + 5 = (13 + 5) + \frac{3}{7} =\]

\[= 18\frac{3}{7}\]

\[\textbf{г)}\ 2\frac{8}{9} + 19 = (2 + 19) + \frac{8}{9} =\]

\[= 21\frac{8}{9}\]

\[\textbf{д)}\ 37\frac{5}{9} - 9 = (37 - 9) + \frac{5}{9} =\]

\[= 28\frac{5}{9}\]

\[\textbf{е)}\ 4\frac{13}{15} - 4 = (4 - 4) + \frac{13}{15} =\]

\[= \frac{13}{15}\]

\[\textbf{ж)}\ 15\frac{1}{10} - 8 =\]

\[= (15 - 8) + \frac{1}{10} = 7\frac{1}{10}\]

\[\textbf{з)}\ 34\frac{5}{11} + 9 = (34 + 9) + \frac{5}{11} =\]

\[= 43\frac{5}{11}\]

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

\[\textbf{в)}\ 3\frac{5}{6} + \frac{1}{6} = 3 + \left( \frac{5}{6} + \frac{1}{6} \right) =\]

\[= 3 + 1 = 4\]

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

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

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

\[\textbf{е)}\ 9\frac{3}{10} + 4\frac{1}{10} = 13\frac{4}{10} = 13\frac{2}{5}\]

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

\[= 18 + 1 = 19\]

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

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

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

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

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

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

\[= (11 - 11) + \left( \frac{1}{2} - \frac{1}{2} \right) = 0\]

\[\textbf{е)}\ 18\frac{4}{9} - 9\frac{3}{9} =\]

\[= (18 - 9) + \left( \frac{4}{9} - \frac{3}{9} \right) = 9\frac{1}{9}\]

\[\textbf{ж)}\ 10\frac{3}{7} - \frac{1}{7} =\]

\[= 10 + \left( \frac{3}{7} - \frac{1}{7} \right) = 10\frac{2}{7}\]

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

\[= (16 - 16) + \left( \frac{3}{10} - \frac{2}{10} \right) =\]

\[= \frac{2}{10} = \frac{1}{5}\]

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

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

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

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

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

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

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

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

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

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

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам