Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 104

\[\boxed{\mathbf{104.}}\]

\[1)\ \frac{y - 16y^{\frac{1}{2}}}{5y^{\frac{1}{4}} + 20} = \frac{y^{\frac{1}{2}} \bullet \left( y^{\frac{1}{2}} - 16 \right)}{5 \bullet \left( y^{\frac{1}{4}} + 4 \right)} =\]

\[= \frac{y^{\frac{1}{2}} \bullet \left( y^{\frac{1}{4}} - 4 \right)\left( y^{\frac{1}{4}} + 4 \right)}{5 \bullet \left( y^{\frac{1}{4}} + 4 \right)} =\]

\[= \frac{y^{\frac{1}{2}} \bullet \left( y^{\frac{1}{4}} - 4 \right)\ }{5} =\]

\[= \text{\ \ }\frac{\sqrt{y} \bullet \left( \sqrt[4]{y} - 4 \right)}{5}\]

\[2)\ \frac{a^{\frac{4}{5}} - b^{\frac{4}{5}}}{a^{\frac{2}{5}} - b^{\frac{2}{5}}} =\]

\[= \frac{\left( a^{\frac{2}{5}} - b^{\frac{2}{5}} \right)\left( a^{\frac{2}{5}} + b^{\frac{2}{5}} \right)}{a^{\frac{2}{5}} - b^{\frac{2}{5}}} =\]

\[= a^{\frac{2}{5}} + b^{\frac{2}{5}} = \ \sqrt[5]{a^{2}} + \sqrt[5]{b^{2}}\]