Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 917

\[1)\log_{|2x + 1|}x^{2} \geq 2\]

\[2\log_{|2x + 1|}|x| \geq 2\]

\[\log_{|2x + 1|}|x| \geq 1\]

\[Основание\ логарифма:\]

\[|2x + 1| < 1\]

\[- 1 < 2x + 1 < 1\]

\[- 2 < 2x < 0\]

\[- 1 < x < 0.\]

\[- 1 < x < 0:\]

\[|x| \leq |2x + 1|\]

\[(2x + 1)^{2} - x^{2} \geq 0\]

\[(2x + 1 - x)(2x + 1 + x) \geq 0\]

\[(x + 1)(3x + 1) \geq 0\]

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

\[x < - 1;\ x > 0:\]

\[|x| \geq |2x + 1|\]

\[(2x + 1)^{2} - x^{2} \leq 0\]

\[(x + 1)(3x + 1) \leq 0\]

\[- 1 \leq x \leq - \frac{1}{3}.\]

\[Область\ определения:\]

\[|2x + 1| > 0\]

\[2x + 1 \neq 0\]

\[2x \neq - 1\]

\[x \neq - \frac{1}{2}.\]

\[Ответ:\ \ x \in \left\lbrack - \frac{1}{3};\ 0 \right).\]

\[2)\log_{x^{2}}|3x + 1| < \frac{1}{2}\]

\[\frac{1}{2}\log_{|x|}|3x + 1| < \frac{1}{2}\]

\[\log_{|x|}|3x + 1| < 1\]

\[|x| < 1:\]

\[|3x + 1| > |x|\]

\[(3x + 1)^{2} - x^{2} > 0\]

\[(3x + 1 - x)(3x + 1 + x) > 0\]

\[(2x + 1)(4x + 1) > 0\]

\[x < - \frac{1}{2};\ \ \ x > - \frac{1}{4}.\]

\[|x| > 1:\]

\[|3x + 1| < |x|\]

\[(3x + 1)^{2} - x^{2} < 0\]

\[(2x + 1)(4x + 1) < 0\]

\[- \frac{1}{2} < x < - \frac{1}{4}.\]

\[Область\ определения:\]

\[|x| \neq 0;\ \ \ x \neq 0.\]

\[Ответ:\ \ \]

\[x \in \left( - 1;\ - \frac{1}{2} \right) \cup \left( - \frac{1}{4};\ 0 \right) \cup (0;\ 1).\]