Задание 497

Приведите к общему знаменателю дроби:

\({\largeа)}\ \frac{2}{3}\) и \(\frac{4}{5};\)
\({\largeб)}\ \frac{3}{4}\) и \(\frac{6}{7};\)
\({\largeв)}\ \frac{8}{9}\) и \(\frac{5}{-9};\)

\({\largeг)}\ \frac{4}{5}\) и \(\frac{3}{-7};\)
\({\largeд)}\ \frac{2}{3}\) и \(\frac{5}{6};\)
\({\largeе)}\ \frac{13}{14}\) и \(\frac{6}{7};\)

\({\largeж)}\ \frac{7}{9}\) и \(\frac{5}{-3};\)
\({\largeз)}\ \frac{1}{5}\) и \(\frac{3}{-10};\)
\({\largeи)}\ \frac{3}{10}\) и \(\frac{4}{15};\)

\({\largeк)}\ \frac{5}{12}\) и \(\frac{1}{16};\)
\({\largeл)}\ \frac{8}{14}\) и \(\frac{5}{-21};\)
\({\largeм)}\ \frac{7}{24}\) и \(\frac{1}{-18}.\)

Источник заимствования: Алгебра. 7 класс. Учебник для общеобразовательных организаций / С.М. Никольский, М.К. Потапов, Н.Н. Решетников, А.В. Шевкин – Просвещение, 2013. – 129 c. ISBN 978-5-09-027739-6

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Решение:

\({\largeа)}\ \frac{2^{\backslash5}}{3\phantom{^{\backslash5}}}=\frac{2\cdot5}{3\cdot5}=\frac{10}{15}\) и \(\frac{4^{\backslash3}}{5\phantom{^{\backslash3}}}=\frac{4\cdot3}{5\cdot3}=\frac{12}{15};\)
\({\largeб)}\ \frac{3^{\backslash7}}{4\phantom{^{\backslash7}}}=\frac{3\cdot7}{4\cdot7}=\frac{21}{28}\) и \(\frac{6^{\backslash4}}{7\phantom{^{\backslash4}}}=\frac{6\cdot4}{7\cdot4}=\frac{24}{28};\)
\({\largeв)}\ \frac{8^{\backslash1}}{9\phantom{^{\backslash1}}}=\frac{8\cdot1}{9\cdot1}=\frac{8}{9}\) и \(\frac{5^{\backslash1}}{-9\phantom{^{\backslash1}}}={-}\frac{5\cdot1}{9\cdot1}={-}\frac{5}{9};\)
\({\largeг)}\ \frac{4^{\backslash7}}{5\phantom{^{\backslash7}}}=\frac{4\cdot7}{5\cdot7}=\frac{28}{35}\) и \(\frac{3^{\backslash5}}{-7\phantom{^{\backslash5}}}={-}\frac{3\cdot5}{7\cdot5}={-}\frac{15}{35};\)
\({\largeд)}\ \frac{2^{\backslash2}}{3\phantom{^{\backslash2}}}=\frac{2\cdot2}{3\cdot2}=\frac{4}{6}\) и \(\frac{5^{\backslash1}}{6\phantom{^{\backslash1}}}=\frac{5\cdot1}{6\cdot1}=\frac{5}{6};\)
\({\largeе)}\ \frac{13^{\backslash1}}{14\phantom{^{\backslash1}}}=\frac{13\cdot1}{14\cdot1}=\frac{13}{14}\) и \(\frac{6^{\backslash2}}{7\phantom{^{\backslash2}}}=\frac{6\cdot2}{7\cdot2}=\frac{12}{14};\)
\({\largeж)}\ \frac{7^{\backslash1}}{9\phantom{^{\backslash1}}}=\frac{7\cdot1}{9\cdot1}=\frac{7}{9}\) и \(\frac{5^{\backslash3}}{-3\phantom{^{\backslash3}}}={-}\frac{5\cdot3}{3\cdot3}=\frac{15}{9};\)
\({\largeз)}\ \frac{1^{\backslash2}}{5\phantom{^{\backslash2}}}=\frac{1\cdot2}{5\cdot2}=\frac{2}{10}\) и \(\frac{3^{\backslash1}}{-10\phantom{^{\backslash1}}}={-}\frac{3\cdot1}{10\cdot1}={-}\frac{3}{10};\)
\({\largeи)}\ \frac{3^{\backslash3}}{10\phantom{^{\backslash3}}}=\frac{3\cdot3}{10\cdot3}=\frac{9}{30}\) и \(\frac{4^{\backslash2}}{15\phantom{^{\backslash2}}}=\frac{4\cdot2}{15\cdot2}=\frac{8}{30};\)
\({\largeк)}\ \frac{5^{\backslash4}}{12\phantom{^{\backslash4}}}=\frac{5\cdot4}{12\cdot4}=\frac{20}{48}\) и \(\frac{1^{\backslash3}}{16\phantom{^{\backslash3}}}=\frac{1\cdot3}{16\cdot3}=\frac{3}{48};\)
\({\largeл)}\ \frac{8^{\backslash3}}{14\phantom{^{\backslash3}}}=\frac{8\cdot3}{14\cdot3}=\frac{24}{42}\) и \(\frac{5^{\backslash2}}{-21\phantom{^{\backslash2}}}={-}\frac{5\cdot2}{21\cdot2}={-}\frac{10}{42};\)
\({\largeм)}\ \frac{7^{\backslash3}}{24\phantom{^{\backslash3}}}=\frac{7\cdot3}{24\cdot3}=\frac{21}{72}\) и \(\frac{1^{\backslash4}}{-18\phantom{^{\backslash4}}}={-}\frac{1\cdot4}{18\cdot4}={-}\frac{4}{72}.\)

Задание 496Задание 498