Вопрос:

17.2. Виконайте віднімання дробів: 1) \(\frac{4d+7}{7d} - \frac{d-6}{6d}\) 2) \(\frac{m-n}{mn} - \frac{p-n}{np}\) 3) \(\frac{6a+2}{ab} - \frac{2a+4}{a^2b}\) 4) \(\frac{c^2-16}{c^6} - \frac{c-9}{c^5}\) 5) \(\frac{1}{x^3} - \frac{1+x^2}{x^5}\) 6) \(\frac{1-ab}{abc} - \frac{1-ad}{acd}\)

Ответ:

Рішення:

1) \(\frac{4d+7}{7d} - \frac{d-6}{6d} = \frac{6(4d+7) - 7(d-6)}{42d} = \frac{24d+42 - 7d+42}{42d} = \frac{17d+84}{42d}\)

2) \(\frac{m-n}{mn} - \frac{p-n}{np} = \frac{p(m-n) - m(p-n)}{mnp} = \frac{pm-pn - mp+mn}{mnp} = \frac{mn-pn}{mnp}\)

3) \(\frac{6a+2}{ab} - \frac{2a+4}{a^2b} = \frac{a(6a+2) - (2a+4)}{a^2b} = \frac{6a^2+2a - 2a-4}{a^2b} = \frac{6a^2-4}{a^2b}\)

4) \(\frac{c^2-16}{c^6} - \frac{c-9}{c^5} = \frac{c^2-16 - c(c-9)}{c^6} = \frac{c^2-16 - c^2+9c}{c^6} = \frac{9c-16}{c^6}\)

5) \(\frac{1}{x^3} - \frac{1+x^2}{x^5} = \frac{x^2 - (1+x^2)}{x^5} = \frac{x^2 - 1 - x^2}{x^5} = \frac{-1}{x^5}\)

6) \(\frac{1-ab}{abc} - \frac{1-ad}{acd} = \frac{d(1-ab) - b(1-ad)}{abcd} = \frac{d-abd - b+abd}{abcd} = \frac{d-b}{abcd}\)

Відповідь:

  • 1) \(\frac{17d+84}{42d}\)
  • 2) \(\frac{mn-pn}{mnp}\)
  • 3) \(\frac{6a^2-4}{a^2b}\)
  • 4) \(\frac{9c-16}{c^6}\)
  • 5) \(\frac{-1}{x^5}\)
  • 6) \(\frac{d-b}{abcd}\)
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