ГДЗ по алгебре 8 класс Макарычев Задание 436

\[\boxed{\text{436\ (436).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x}{x + \sqrt{y}} = \frac{x\left( x - \sqrt{y} \right)}{\left( x + \sqrt{y} \right)\left( x - \sqrt{y} \right)} =\]

\[= \frac{x(x - \sqrt{y})}{x^{2} - y}\]

\[\textbf{б)}\ \frac{b}{a - \sqrt{b}} = \frac{b\left( a + \sqrt{b} \right)}{\left( a - \sqrt{b} \right)\left( a + \sqrt{b} \right)} =\]

\[= \frac{b(a + \sqrt{b})}{a^{2} - b}\]

\[\textbf{в)}\ \frac{4}{\sqrt{10} - \sqrt{2}} =\]

\[= \frac{4\left( \sqrt{10} + \sqrt{2} \right)}{\left( \sqrt{10} - \sqrt{2} \right)\left( \sqrt{10} + \sqrt{2} \right)} =\]

\[= \frac{4\left( \sqrt{10} + \sqrt{2} \right)}{10 - 2} =\]

\[= \frac{4\left( \sqrt{10} + \sqrt{2} \right)}{8} =\]

\[= \frac{\sqrt{10} + \sqrt{}2}{2}\]

\[\textbf{г)}\ \frac{12}{\sqrt{3} + \sqrt{6}} =\]

\[= \frac{12\left( \sqrt{3} - \sqrt{6} \right)}{\left( \sqrt{3} + \sqrt{6} \right)\left( \sqrt{3} - \sqrt{6} \right)} =\]

\[= \frac{12\left( \sqrt{3} - \sqrt{6} \right)}{3 - 6} =\]

\[= \frac{12\left( \sqrt{3} - \sqrt{6} \right)}{- 3} =\]

\[= - 4\left( \sqrt{3} - \sqrt{6} \right) = 4\sqrt{6} - 4\sqrt{3}\]

\[\textbf{д)}\ \frac{9}{3 - 2\sqrt{2}} =\]

\[= \frac{9\left( 3 + 2\sqrt{2} \right)}{\left( 3 - 2\sqrt{2} \right)\left( 3 + 2\sqrt{2} \right)} =\]

\[= \frac{9\left( 3 + 2\sqrt{2} \right)}{9 - 8} = \frac{27 + 18\sqrt{2}}{1} =\]

\[= 27 + 18\sqrt{2}\]

\[\textbf{е)}\ \frac{14}{1 + 5\sqrt{2}} =\]

\[= \frac{14\left( 1 - 5\sqrt{2} \right)}{\left( 1 + 5\sqrt{2} \right)\left( 1 - 5\sqrt{2} \right)} =\]

\[= \frac{14\left( 1 - 5\sqrt{2} \right)}{1 - 25 \cdot 2} = \frac{14\left( 1 - 5\sqrt{2} \right)}{1 - 50} =\]

\[= - \frac{14\left( 1 - 5\sqrt{2} \right)}{49} =\]

\[= - \frac{2\left( 1 - 5\sqrt{2} \right)}{7} = \frac{10\sqrt{2} - 2}{7}\]