Решебник по математике 4 класс Петерсон рабочая тетрадь Часть 1, 2, 3 Часть 2 | Страница 14

\[\boxed{Страница\ 15.}\]

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

\[1 < \frac{5}{4} < 2\]

\[2 < \frac{9}{4} < 3\]

\[3 < \frac{15}{4} < 4\]

\[4 < \frac{18}{4} < 5\]

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

\[\textbf{а)}\ Мы\ пока\ не\ знаем,\ как\ \]

\[образуются\ смешанные\ числа.\]

\[\textbf{б)}\]

\[\frac{7}{3} = 2 + \frac{1}{3}\]

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

\[\textbf{а)}\ 2 + \frac{4}{5} = 2\frac{4}{5}\]

\[\textbf{б)}\ 9 + \frac{3}{16} = 9\frac{3}{16}\]

\[\textbf{в)}\ 15 + \frac{1}{2} = 15\frac{1}{2}\]

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

\[\textbf{а)}\ 1\frac{6}{7} = 1 + \frac{6}{7}\]

\[\textbf{б)}\ 3\frac{2}{9} = 3 + \frac{2}{9}\]

\[\textbf{в)}\ 5\frac{8}{10} = 5 + \frac{8}{10}\]

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

\[\textbf{а)}\ 243\ :75 = 3\ (ост.\ 18)\]

\[Проверка:\]

\[3 \cdot 75 + 18 = 225 + 18 = 243\]

\[\textbf{б)}\ 6091\ :564 = 10\ (ост.\ 451)\]

\[Проверка:\]

\[564 \cdot 10 + 451 = 5640 + 451 =\]

\[= 6091\]

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

\[1\frac{1}{2};\ \ \ 2\frac{1}{3};\ \ \ 3\frac{1}{4};\ \ \ 4\frac{1}{5};\ \ \ 5\frac{1}{6};\ \ \]

\[\ 6\frac{1}{7};\ \ \ 7\frac{1}{8}.\]