Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 293

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\[\textbf{а)}\ \frac{57}{4} = 14\frac{1}{4};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\frac{10}{3} = 3\frac{1}{3};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\frac{17}{11} = 1\frac{6}{11};\ \]

\[\frac{666}{100} = 6\frac{66}{100};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\frac{1477}{211} = 7;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\frac{65\ 070}{10\ 000} = 6\frac{5070}{10000}\text{.\ \ }\]

\[\textbf{б)}\ 1\frac{3}{4} = \frac{7}{4};\ \]

\[6\frac{4}{15} = \frac{15 \cdot 6 + 4}{15} = \frac{94}{15};\ \]

\[9\frac{11}{19} = \frac{19 \cdot 9 + 11}{19} = \frac{171 + 11}{19} =\]

\[= \frac{182}{19};\ \]

\[10\frac{53}{61} = \frac{61 \cdot 10 + 53}{61} =\]

\[= \frac{610 + 53}{61} = \frac{663}{61};\ \]

\[18\frac{1}{100} = \frac{100 \cdot 18 + 1}{100} = \frac{1\ 801}{100};\]

\[7\frac{15}{10\ 000} = \frac{10\ 000 \cdot 7 + 15}{10\ 000} =\]

\[= \frac{70\ 015}{10\ 000}.\]