Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1614

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

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

\[\sqrt{(2x + 1)^{2}} > 1\]

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

\[4x^{2} + 4x + 1 > 1\]

\[4x^{2} + 4x > 0\]

\[x^{2} + x > 0\]

\[(x + 1) \bullet x > 0\]

\[x < - 1\ \ и\ \ x > 0.\]

\[x < - 1\ или\ x > 0:\]

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

\[x^{2} \geq 4x^{2} + 4x + 1\]

\[3x^{2} + 4x + 1 \leq 0\]

\[D = 16 - 12 = 4\]

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

\[x_{2} = \frac{- 4 + 2}{2 \bullet 3} = - \frac{2}{6} = - \frac{1}{3};\]

\[(x + 1)\left( x + \frac{1}{3} \right) \leq 0\]

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

\[- 1 < x < 0:\]

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

\[(x + 1)\left( x + \frac{1}{3} \right) \geq 0\]

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

\[Ответ:\ \ - \frac{1}{3} \leq x < 0.\]

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

\[2 \bullet \log_{x^{2}}\sqrt{(3x + 1)^{2}} < 1\]

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

\[x^{2} > 1\]

\[x < - 1\ и\ x > 1.\]

\[x < - 1\ или\ x > 1:\]

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

\[9x^{2} + 6x + 1 < x^{2}\]

\[8x^{2} + 6x + 1 < 0\]

\[D = 36 - 32 = 4\]

\[x_{1} = \frac{- 6 - 2}{2 \bullet 8} = - \frac{1}{2};\]

\[x_{2} = \frac{- 6 + 2}{2 \bullet 8} = - \frac{1}{4};\]

\[\left( x + \frac{1}{2} \right)\left( x + \frac{1}{4} \right) < 0\]

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

\[- 1 < x < 1:\]

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

\[\left( x + \frac{1}{2} \right)\left( x + \frac{1}{4} \right) > 0\]

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

\[Имеет\ смысл\ при:\]

\[x \neq 0\ \ и\ \ x \neq \pm 1.\]

\[Ответ:\ \ - 1 < x < - \frac{1}{2};\ \ \]

\[- \frac{1}{4} < x < 0;\ \ 0 < x < 1.\]