Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 613

\[\boxed{\mathbf{613}.}\]

\[1)\ \sqrt[3]{x^{3} - 2} = x - 2;\]

\[x^{3} - 2 = (x - 2)^{3};\]

\[x^{3} - 2 = x^{3} - 3x^{2} \bullet 2 +\]

\[+ 3x \bullet 2^{2} - 2^{3};\]

\[x^{3} - 2 = x^{3} - 6x^{2} + 12x - 8;\]

\[6x^{2} - 12x + 6 = 0;\]

\[x^{2} - 2x + 1 = 0;\]

\[(x - 1)^{2} = 0;\]

\[x - 1 = 0;\]

\[x = 1;\]

\[Ответ:\ \ x = 1.\]

\[2)\ \sqrt[3]{x^{3} - 5x^{2} + 16x - 5} =\]

\[= x - 2;\]

\[x^{3} - 5x^{2} + 16x - 5 = (x - 2)^{3};\]

\[x^{3} - 5x^{2} + 16x - 5 =\]

\[= x^{3} - 3x^{2} \bullet 2 + 3x \bullet 2^{2} - 2^{3};\]

\[x^{3} - 5x^{2} + 16x - 5 = x^{3} -\]

\[- 6x^{2} + 12x - 8;\]

\[x^{2} + 4x + 3 = 0;\]

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

\[x_{1} = \frac{- 4 - 2}{2} = - 3\ \ и\ \]

\[\ x_{2} = \frac{- 4 + 2}{2} = - 1;\]

\[Ответ:\ \ x_{1} = - 3;\ \ x_{2} = - 1.\]