Страница 224 номер 814, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Упростите выражение: 1) \(\frac{\sqrt{a} - \sqrt{b}}{\sqrt{ab}} + \frac{\sqrt{a} - \sqrt{c}}{\sqrt{bc}}\); 2) \(\frac{d - 49}{d + 12\sqrt{d} + 36} \cdot \frac{4\sqrt{d} + 24}{3\sqrt{d} + 21}\).
1)
\(\dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{ab}} + \dfrac{\sqrt{a} - \sqrt{c}}{\sqrt{bc}} = \dfrac{\sqrt{bc}\left( \sqrt{a} - \sqrt{b} \right) + \sqrt{ab}\left( \sqrt{a} - \sqrt{c} \right)}{\sqrt{ab} \cdot \sqrt{bc}} =\)
\(= \dfrac{\sqrt{abc} - b\sqrt{c} + a\sqrt{b} - \sqrt{abc}}{b\sqrt{ac}} = \dfrac{\sqrt{b}\left( a - \sqrt{bc} \right)}{b\sqrt{ac}} = \dfrac{a - \sqrt{bc}}{\sqrt{abc}}\)
2)
\(\dfrac{d - 49}{d + 12\sqrt{d} + 36} \cdot \dfrac{4\sqrt{d} + 24}{3\sqrt{d} + 21} =\)
\(= \dfrac{\left( \sqrt{d} - 7 \right)(\sqrt{d} + 7)}{\left( \sqrt{d} + 6 \right)^{2}} \cdot \dfrac{4(\sqrt{d} + 6)}{3(\sqrt{d} + 7)} = \dfrac{4(\sqrt{d} - 7)}{3(\sqrt{d} + 6)} = \dfrac{4\sqrt{d} - 28}{3\sqrt{d} + 18}\)
Ответ: 1) \(\frac{a - \sqrt{bc}}{\sqrt{abc}}\); 2) \(\frac{4\sqrt{d} - 28}{3\sqrt{d} + 18}\).
