Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 893

\[1)\ |x - 1|\left( x^{4} - 2x^{2} - 3 \right) \geq 0\]

\[D = 4 + 12 = 16\]

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

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

\[|x - 1|\left( x^{2} + 1 \right)\left( x^{2} - 3 \right) \geq 0\]

\[\left( x + \sqrt{3} \right)\left( x - \sqrt{3} \right) \geq 0\text{\ \ }\]

\[x - 1 = 0\]

\[x \leq - \sqrt{3};\ \ x \geq \sqrt{3};\ \ \ x = 1.\]

\[Ответ:\ \ \]

\[x \in \left( - \infty;\ - \sqrt{3} \right\rbrack \cup \left\{ 1 \right\} \cup \left\lbrack \sqrt{3};\ + \infty \right).\]

\[2)\ \left| x^{2} - 9 \right|\left( x^{4} - 2x^{2} - 8 \right) \leq 0\]

\[D = 4 + 32 = 36\]

\[x_{1}^{2} = \frac{2 - 6}{2} = - 2;\]

\[x_{2}^{2} = \frac{2 + 6}{2} = 4;\]

\[\left| x^{2} - 9 \right|\left( x^{2} + 2 \right)\left( x^{2} - 4 \right) \geq 0\]

\[(x + 2)(x - 2) \leq 0\]

\[- 2 \leq x \leq 2.\text{\ \ \ }\]

\[x^{2} - 9 = 0\]

\[x = \pm 3.\]

\[Ответ:\ \ x \in \lbrack - 2;\ 2\rbrack \cup \left\{ - 3;\ 3 \right\}.\]