Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 318

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\[\textbf{а)}\ \frac{1}{2} = \frac{3}{6} = \frac{12}{24}\]

\[\frac{1}{2} = \frac{1 \cdot 3}{2 \cdot 3} = \frac{3}{6};\ \ \ \]

\[\text{\ \ }\frac{1}{2} = \frac{1 \cdot 12}{2 \cdot 12} = \frac{12}{24}.\]

\[\textbf{б)}\ \frac{4}{6} = \frac{12}{18} = \frac{20}{30}\ \]

\[\frac{4}{6} = \frac{4 \cdot 3}{6 \cdot 3} = \frac{12}{18};\ \]

\[\text{\ \ \ }\frac{4}{6} = \frac{4 \cdot 5}{6 \cdot 5} = \frac{20}{30}\text{.\ }\]

\[\textbf{в)}\ \frac{3}{5} = \frac{15}{25} = \frac{27}{45}\ \]

\[\frac{3}{5} = \frac{3 \cdot 5}{5 \cdot 5} = \frac{15}{25};\ \ \ \]

\[\text{\ \ }\frac{3}{5} = \frac{3 \cdot 9}{5 \cdot 9} = \frac{27}{45}.\]

\[\textbf{г)}\ \frac{12}{14} = \frac{6}{7} = \frac{60}{70}\ \]

\[\frac{6}{7} = \frac{6 \cdot 2}{7 \cdot 2} = \frac{12}{14};\ \ \]

\[\text{\ \ }\frac{12}{14} = \frac{12 \cdot 5}{14 \cdot 5} = \frac{60}{70}\text{.\ \ }\]