Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 130

\[\boxed{\mathbf{130.}}\]

\[1)\ y = \sqrt[5]{x}\text{\ \ }и\ \ y = x^{\frac{3}{5}}\]

\[\sqrt[5]{x} = x^{\frac{3}{5}}\]

\[x^{\frac{1}{5}} - x^{\frac{3}{5}} = 0\]

\[x^{\frac{1}{5}} \bullet \left( 1 - x^{\frac{2}{5}} \right) = 0\]

\[x^{\frac{1}{5}} \bullet \left( 1 - x^{\frac{1}{5}} \right)\left( 1 + x^{\frac{1}{5}} \right) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = 1;\ \ \ \ x_{3} = - 1;\]

\[y_{1} = 0;\ \ \ y_{2} = 1;\ \ \ y_{3} = - 1;\]

\[Функция\ y = x^{\frac{3}{5}}\ определена\ \]

\[при\ \ x \geq 0.\]

\[Ответ:\ \ (0;\ 0);\ \ (1;\ 1).\]

\[2)\ y = \sqrt[7]{x}\text{\ \ }и\ \ y = x^{\frac{5}{7}}\]

\[\sqrt[7]{x} = x^{\frac{5}{7}}\]

\[x^{\frac{1}{7}} - x^{\frac{5}{7}} = 0\]

\[x^{\frac{1}{7}} \bullet \left( 1 - x^{\frac{4}{7}} \right) = 0\]

\[x^{\frac{1}{7}} \bullet \left( 1 - x^{\frac{2}{7}} \right)\left( 1 + x^{\frac{2}{7}} \right) = 0\]

\[x^{\frac{1}{7}} \bullet \left( 1 - x^{\frac{1}{7}} \right)\left( 1 + x^{\frac{1}{7}} \right) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = 1;\ \ \ x_{3} = - 1;\]

\[y_{1} = 0;\ \ \ y_{2} = 1;\ \ \ y_{3} = - 1;\]

\[Функция\ y = x^{\frac{5}{7}}\ определена\ \]

\[при\ \ x \geq 0.\]

\[Ответ:\ \ (0;\ 0);\ \ (1;\ 1).\]