Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 45

\[\boxed{\mathbf{45}\mathbf{.}}\]

\[1)\ \sqrt[6]{2x - 3};\]

\[2x - 3 \geq 0;\]

\[2x \geq 3;\]

\[x \geq \frac{3}{2};\]

\[Ответ:\ \ x \geq 1,5.\]

\[2)\ \sqrt[6]{x + 3};\]

\[x + 3 \geq 0;\]

\[x \geq - 3;\]

\[Ответ:\ \ x \geq - 3.\]

\[3)\ \sqrt[6]{2x^{2} - x - 1};\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{1 - 3}{2 \bullet 2} = \frac{- 2}{4} = - \frac{1}{2} = - 0,5;\]

\[x_{2} = \frac{1 + 3}{2 \bullet 2} = \frac{4}{4} = 1;\]

\[(x + 0,5)(x - 1) \geq 0;\]

\[x \leq - 0,5\ \ и\ \ x \geq 1;\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 0,5) \cup (1;\ + \infty).\]

\[4)\ \sqrt[4]{\frac{2 - 3x}{2x - 4}};\]

\[\frac{2 - 3x}{2x - 4} \geq 0;\]

\[(2 - 3x)(2x - 4) \geq 0;\]

\[(3x - 2)(2x - 4) \leq 0;\]

\[\frac{2}{3} \leq x < 2;\]

\[Ответ:\ \ x \in \left( \frac{2}{3};\ 2 \right).\]