Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 31

\[\boxed{\mathbf{31}\mathbf{.}}\]

\[1)\ x^{4} = 256\]

\[x^{4} - 256 = 0\]

\[\left( x^{2} - 16 \right)\left( x^{2} + 16 \right) = 0\]

\[\left( x^{2} - 16 \right) = 0\]

\[(x - 4)(x + 4) = 0\]

\[x_{1} = 4\ \ и\ \ x_{2} = - 4\]

\[Ответ:\ \ x = \pm 4.\]

\[2)\ x^{5} = - \frac{1}{32}\]

\[x = \sqrt[5]{- \frac{1}{32}} = - \sqrt[5]{\left( \frac{1}{2} \right)^{5}} = - \frac{1}{2}\]

\[Ответ:\ \ x = - 0,5.\]

\[3)\ 5x^{5} = - 160\]

\[x^{5} = - 32\]

\[x = \sqrt[5]{- 32} = - \sqrt[5]{2^{5}} = - 2\]

\[Ответ:\ \ x = - 2.\]

\[4)\ 2x^{6} = 128\]

\[x^{6} = 64\]

\[x = \pm \sqrt[6]{64} = \pm \sqrt[6]{2^{6}} = \pm 2\]

\[Ответ:\ \ x = \pm 2.\]