Страница 405 номер 1285, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\left( \frac{a + \sqrt{ab}}{\sqrt{a^{2} + ab}} - \frac{\sqrt{ab + b^{2}}}{\sqrt{ab} + b} \right)^{-2} - \frac{\sqrt{a^{3}b} + \sqrt{ab^{3}}}{2ab} =\)
\(= \left( \dfrac{\sqrt{a}\left( \sqrt{a} + \sqrt{b} \right)}{\sqrt{a} \cdot \sqrt{a + b}} - \dfrac{\sqrt{b} \cdot \sqrt{a + b}}{\sqrt{b}\left( \sqrt{a} + \sqrt{b} \right)} \right)^{-2} - \dfrac{a\sqrt{ab} + b\sqrt{ab}}{2ab} =\)
\(= \left( \dfrac{\left( \sqrt{a} + \sqrt{b} \right)^{2} - (a + b)}{\sqrt{a + b}\left( \sqrt{a} + \sqrt{b} \right)} \right)^{-2} - \dfrac{\sqrt{ab}(a + b)}{2ab} =\)
\(= \left( \dfrac{\sqrt{a + b}\left( \sqrt{a} + \sqrt{b} \right)}{2\sqrt{ab}} \right)^{2} - \dfrac{\sqrt{ab}(a + b)}{2ab} =\)
\(= \dfrac{(a + b)\left( a + 2\sqrt{ab} + b \right)}{4ab} - \dfrac{\sqrt{ab}(a + b)}{2ab} =\)
\(= \dfrac{a^{2} + ab + 2a\sqrt{ab} + 2b\sqrt{ab} + ab + b^{2} - 2a\sqrt{ab} - 2b\sqrt{ab}}{4ab} =\)
\(= \dfrac{a^{2} + 2ab + b^{2}}{4ab} = \dfrac{(a + b)^{2}}{4ab};\)

2) \(\left( \sqrt{a} + \sqrt{b} \right)^{-2} \cdot \left( \frac{1}{a} + \frac{1}{b} \right) + \frac{2\left( a^{-\frac{1}{2}} + b^{-\frac{1}{2}} \right)}{\left( \sqrt{a} + \sqrt{b} \right)^{3}} =\)
\(= \frac{1}{\left( \sqrt{a} + \sqrt{b} \right)^{2}} \cdot \frac{a + b}{ab} + \frac{2\left( \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} \right)}{\left( \sqrt{a} + \sqrt{b} \right)^{3}} =\)
\(= \dfrac{a + b}{ab\left( \sqrt{a} + \sqrt{b} \right)^{2}} + \dfrac{2\left( \sqrt{a} + \sqrt{b} \right)}{\sqrt{ab}\left( \sqrt{a} + \sqrt{b} \right)^{3}} =\)
\(= \dfrac{a + b + 2\sqrt{ab}}{ab\left( \sqrt{a} + \sqrt{b} \right)^{2}} = \dfrac{\left( \sqrt{a} + \sqrt{b} \right)^{2}}{ab\left( \sqrt{a} + \sqrt{b} \right)^{2}} = \dfrac{1}{ab}.\)