ГДЗ по алгебре 8 класс Макарычев Задание 431

\[\boxed{\text{431\ (431).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x}{\sqrt{5}} = \frac{x \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{x\sqrt{5}}{5}\]

\[\textbf{б)}\ \frac{3}{\sqrt{b}} = \frac{3 \cdot \sqrt{b}}{\sqrt{b} \cdot \sqrt{b}} = \frac{3\sqrt{b}}{b}\]

\[\textbf{в)}\ \frac{2}{7\sqrt{y}} = \frac{2 \cdot \sqrt{y}}{7\sqrt{y} \cdot \sqrt{y}} = \frac{2\sqrt{y}}{7y}\]

\[\textbf{г)}\ \frac{a}{b\sqrt{b}} = \frac{a \cdot \sqrt{b}}{b\sqrt{b} \cdot \sqrt{b}} \cdot = \frac{a\sqrt{b}}{b^{2}}\]

\[\textbf{д)}\ \frac{4}{\sqrt{a + b}} = \frac{4 \cdot \sqrt{a + b}}{\sqrt{a + b} \cdot \sqrt{a + b}} =\]

\[= \frac{4\sqrt{a + b}}{a + b}\]

\[\textbf{е)}\ \frac{1}{\sqrt{a - b}} = \frac{1 \cdot \sqrt{a - b}}{\sqrt{a - b} \cdot \sqrt{a - b}} =\]

\[= \frac{\sqrt{a - b}}{a - b}\]

\[\textbf{ж)}\ \frac{5}{2\sqrt{3}} = \frac{5 \cdot \sqrt{3}}{2\sqrt{3} \cdot \sqrt{3}} = \frac{5\sqrt{3}}{2 \cdot 3} =\]

\[= \frac{5\sqrt{3}}{6}\]

\[\textbf{з)}\ \frac{8}{3\sqrt{2}} = \frac{8 \cdot \sqrt{2}}{3\sqrt{2} \cdot \sqrt{2}} = \frac{8\sqrt{2}}{3 \cdot 2} =\]

\[= \frac{4\sqrt{2}}{3}\]

\[\textbf{и)}\ \frac{3\sqrt{5}}{5\sqrt{2}} = \frac{3\sqrt{5} \cdot \sqrt{2}}{5\sqrt{2} \cdot \sqrt{2}} = \frac{3\sqrt{10}}{5 \cdot 2} =\]

\[= \frac{3\sqrt{10}}{10}\]