Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 253

\[\boxed{\text{5.253.}}\]

\[\textbf{а)}\ 8\frac{1}{2} = \frac{2 \cdot 8 + 1}{2} = \frac{17}{2}\ \]

\[6\frac{3}{4} = \frac{4 \cdot 6 + 3}{4} = \frac{27}{4}\ \]

\[5\frac{5}{9} = \frac{9 \cdot 5 + 5}{9} = \frac{50}{9}\ \]

\[2\frac{7}{10} = \frac{10 \cdot 2 + 7}{10} = \frac{27}{10}\ \]

\[\textbf{б)}\ 3 = \frac{3 \cdot 6}{6} = \frac{18}{6}\text{\ \ \ \ \ \ \ \ \ }\]

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

\[5 = \frac{5 \cdot 6}{6} = \frac{30}{6}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[5 = \frac{5 \cdot 3}{3} = \frac{15}{3}\ \]

\[7 = \frac{7 \cdot 6}{6} = \frac{42}{6}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[7 = \frac{7 \cdot 3}{3} = \frac{21}{3}\ \]

\[34 = \frac{34 \cdot 6}{6} = \frac{204}{6}\text{\ \ \ \ \ \ \ }\]

\[34 = \frac{34 \cdot 3}{3} = \frac{102}{3}\ \]