Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1124

\[\boxed{\mathbf{1124}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{a^{3}}{\sqrt{b}} = \frac{a^{3}\sqrt{b}}{\sqrt{b} \cdot \sqrt{b}} = \frac{a^{3}\sqrt{b}}{b}\]

\[2)\ \frac{7}{a\sqrt{a}} = \frac{7\sqrt{a}}{a\sqrt{a} \cdot \sqrt{a}} = \frac{7\sqrt{a}}{a^{2}}\]

\[3)\ \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{\sqrt{13} \cdot \sqrt{13}} = \frac{2\sqrt{13}}{13}\]

\[4)\ \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\]

\[\ 5)\frac{n + 9}{\sqrt{n + 9}} = \frac{(n + 9)\sqrt{n + 9}}{\sqrt{n + 9} \cdot \sqrt{n + 9}} =\]

\[= \frac{(n + 9)\sqrt{n + 9}}{(n + 9)} = \sqrt{n + 9}\]

\[6)\ \frac{3}{\sqrt{13} - 2} =\]

\[= \frac{3 \cdot \left( \sqrt{13} + 2 \right)}{\left( \sqrt{13} - 2 \right)\left( \sqrt{13} + 2 \right)} =\]

\[= \frac{3 \cdot (\sqrt{13} + 2)}{13 - 4} = \frac{\sqrt{13} + 2}{3}\]

\[7)\ \frac{6}{\sqrt{21} + \sqrt{15}} =\]

\[= \frac{6 \cdot \left( \sqrt{21} - \sqrt{15} \right)}{\left( \sqrt{21} + \sqrt{15} \right)\left( \sqrt{21} - \sqrt{15} \right)} =\]

\[= \frac{6 \cdot (\sqrt{21} - \sqrt{15})}{21 - 15} = \sqrt{21} - \sqrt{15}\]

\[8)\ \frac{18}{\sqrt{47} - \sqrt{29}\ } =\]

\[= \frac{18 \cdot \left( \sqrt{47} + \sqrt{29}\ \right)}{\left( \sqrt{47} - \sqrt{29}\ \right)\left( \sqrt{47} + \sqrt{29}\ \right)} =\]

\[= \frac{18 \cdot \left( \sqrt{47} + \sqrt{29}\ \right)}{47 - 29} =\]

\[= \sqrt{47} + \sqrt{29}\ \]