Решебник по математике 6 класс Ерина Виленкин рабочая тетрадь Часть 1, 2 Часть 1 | Страница 48

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\[10.\ Сравнение,\ сложение\ и\ \]

\[вычитание\ обыкновенных\ \]

\[дробей\]

\[\boxed{\mathbf{1.}}\]

\[1)\ к\ общему\ знаменателю.\]

\[\boxed{\mathbf{2.}}\]

\[\textbf{а)}\frac{1}{5} < \frac{3}{10}\]

\[\frac{1}{2} < \frac{3}{4}\]

\[\frac{7}{15} > \frac{2}{5}\]

\[\frac{2}{9} = \frac{6}{27}\text{\ \ }\]

\[\textbf{б)}\frac{3}{5} > \frac{21}{40}\]

\[\frac{7}{12} < \frac{39}{48}\]

\[\frac{1}{17} = \frac{3}{51}\]

\[\frac{3}{8} < \frac{5}{6}\]

\[\textbf{в)}\frac{1}{7} > \frac{1}{15}\]

\[\frac{5}{13} < \frac{5}{11}\]

\[\frac{17}{40} < \frac{17}{21}\]

\[\frac{5}{11} > \frac{5}{18}\]

\[\textbf{г)}\ 2 = \frac{4}{2}\]

\[\frac{5}{18} < \frac{7}{15}\]

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

\[3 > \frac{7}{7}\]

\[\boxed{\mathbf{3.}}\]

\[\textbf{а)}\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}\]

\[\textbf{б)}\frac{2}{15} + \frac{7}{10} = \frac{4}{30} + \frac{21}{30} = \frac{25}{30}\]

\[\textbf{в)}\frac{3}{16} + \frac{9}{20} = \frac{15}{80} + \frac{36}{80} = \frac{51}{80}\]

\[\textbf{г)}\frac{5}{8} + \frac{11}{12} = \frac{15}{24} + \frac{22}{24} = \frac{37}{24}\]

\[\textbf{д)}\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}\]

\[\textbf{е)}\frac{11}{36} - \frac{5}{18} = \frac{11}{36} - \frac{10}{36} = \frac{1}{36}\]

\[\textbf{ж)}\frac{9}{10} - \frac{7}{8} = \frac{36}{40} - \frac{35}{40} = \frac{1}{40}\]

\[\textbf{з)}\frac{7}{22} - \frac{3}{55} = \frac{35}{110} - \frac{6}{110} = \frac{29}{110}\]