Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Обыкновенные дроби Задание 170

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\[\textbf{а)}\ \frac{4}{18} + \text{n\ }\]

\[n = \frac{1}{18}:\ \ \ \ \ \ \ \ \]

\[\frac{4}{18} + \frac{1}{18} = \frac{4 + 1}{18} = \frac{5}{18};\]

\[n = \frac{3}{18}:\ \ \ \ \ \ \ \ \]

\[\frac{4}{18} + \frac{3}{18} = \frac{4 + 3}{18} = \frac{7}{18};\ \]

\[n = \frac{6}{18}:\ \ \ \ \ \ \ \ \]

\[\frac{4}{18} + \frac{6}{18} = \frac{4 + 6}{18} = \frac{10}{18}\text{.\ }\]

\[\textbf{б)}\ m - \frac{1}{9}\]

\[m = \frac{6}{9}:\ \ \ \ \ \ \ \ \ \ \]

\[\frac{6}{9} - \frac{1}{9} = \frac{6 - 1}{9} = \frac{5}{9};\]

\[m = \frac{5}{9}:\ \ \ \ \ \ \ \ \ \ \]

\[\frac{6}{9} - \frac{5}{9} = \frac{6 - 5}{9} = \frac{1}{9};\ \]

\[m = \frac{2}{9}:\ \ \ \ \ \ \ \ \ \ \]

\[\frac{6}{9} - \frac{2}{9} = \frac{6 - 2}{9} = \frac{4}{9}\text{.\ }\]

\[\textbf{в)}\ \frac{4}{15} + \frac{3}{15} + x = \frac{4 + 3}{15} + x =\]

\[= \frac{7}{15} + x\]

\[x = \frac{1}{15}:\ \ \ \ \ \ \ \ \ \]

\[\frac{7}{15} + \frac{1}{15} = \frac{7 + 1}{15} = \frac{8}{15};\ \]

\[x = \frac{3}{15}:\ \ \ \ \ \ \ \ \ \]

\[\frac{7}{15} + \frac{3}{15} = \frac{7 + 3}{15} = \frac{10}{15};\ \]

\[x = \frac{6}{15}:\ \ \ \ \ \ \ \ \ \]

\[\frac{7}{15} + \frac{6}{15} = \frac{7 + 6}{15} = \frac{13}{15}\text{.\ }\]

\[\textbf{г)}\ \frac{13}{19} - \frac{3}{19} - z = \frac{13 - 3}{19} - z =\]

\[= \frac{10}{19} - z\ \]

\[z = \frac{3}{19}:\ \ \ \ \ \ \ \ \ \]

\[\frac{10}{19} - \frac{3}{19} = \frac{10 - 3}{19} = \frac{7}{19};\]

\[z = \frac{5}{19}:\ \ \ \ \ \ \ \ \ \]

\[\frac{10}{19} - \frac{5}{19} = \frac{10 - 5}{19} = \frac{5}{19};\ \]

\[z = \frac{7}{19}:\ \ \ \ \ \ \ \ \ \]

\[\frac{10}{19} - \frac{7}{19} = \frac{10 - 7}{19} = \frac{3}{19}\text{.\ \ }\]