ГДЗ по математике 5 класс Виленкин ФГОС Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 488

\[\boxed{\text{5.488.}}\]

\[\textbf{а)}\ \frac{3}{7}\text{a\ }\]

\[a = \frac{3}{7}:\]

\[\frac{3}{7} \cdot \frac{3}{7} = \frac{9}{49}.\]

\[a = \frac{119}{66}:\]

\[\frac{3}{7} \cdot \frac{119}{66} = \frac{3 \cdot 119}{7 \cdot 66} = \frac{1 \cdot 17}{1 \cdot 22} = \frac{17}{22}.\]

\[a = \frac{28}{33}:\]

\[\frac{3}{7} \cdot \frac{28}{33} = \frac{3 \cdot 28}{7 \cdot 33} = \frac{1 \cdot 4}{1 \cdot 11} = \frac{4}{11}.\]

\[a = 1:\]

\[\frac{3}{7} \cdot 1 = \frac{3}{7}.\]

\[\textbf{б)}\ \frac{5}{12}b\]

\[b = \frac{1}{5}:\]

\[\frac{5}{12} \cdot \frac{1}{5} = \frac{1}{12}.\]

\[b = \frac{5}{12}:\]

\[\frac{5}{12} \cdot \frac{5}{12} = \frac{25}{144}.\]

\[b = \frac{6}{5}:\]

\[\frac{5}{12} \cdot \frac{6}{5} = \frac{6}{12} = \frac{1}{2}.\]

\[b = \frac{84}{25}:\]

\[\frac{5}{12} \cdot \frac{84}{25} = \frac{5 \cdot 84}{12 \cdot 25} = \frac{1 \cdot 7}{1 \cdot 5} = \frac{7}{5} =\]

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

\[b = 0:\]

\[\frac{5}{12} \cdot 0 = 0.\]