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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1412
Задание 1412
\[\boxed{\mathbf{1412}\mathbf{.}}\]
\[1)\log_{\frac{1}{2}}\left( \log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \right) \leq 0\]
\[\log_{\frac{1}{2}}\left( \log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \right) \leq \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{0}\]
\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \geq 1\]
\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \geq \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{1}\]
\[\frac{3x + 1}{x - 1} \leq \frac{1}{2}\ \ \ \ \ | \bullet 2(x - 1)^{2}\]
\[2(3x + 1)(x - 1) \leq (x - 1)^{2}\]
\[2\left( 3x^{2} - 3x + x - 1 \right) \leq x^{2} - 2x + 1\]
\[6x^{2} - 4x - 2 \leq x^{2} - 2x + 1\]
\[5x^{2} - 2x - 3 \leq 0\]
\[D = 4 + 60 = 64\]
\[x_{1} = \frac{2 - 8}{2 \bullet 5} = - \frac{3}{5};\]
\[x_{2} = \frac{2 + 8}{2 \bullet 5} = 1;\]
\[\left( x + \frac{3}{5} \right)(x - 1) \leq 0\]
\[- \frac{3}{5} \leq x \leq 1.\]
\[Имеет\ смысл\ при:\]
\[\frac{3x + 1}{x - 1} > 0\]
\[(3x + 1)(x - 1) > 0\]
\[x < - \frac{1}{3}\text{\ \ }и\ \ x > 1.\]
\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} > 0\]
\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} > \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{0}\]
\[\frac{3x + 1}{x - 1} < 1\ \ \ \ \ | \bullet (x - 1)^{2}\]
\[(3x + 1)(x - 1) < (x - 1)^{2}\]
\[3x^{2} - 3x + x - 1 < x^{2} - 2x + 1\]
\[2x^{2} < 2\]
\[x < 1\]
\[- 1 < x < 1.\]
\[Ответ:\ \ - \frac{3}{5} \leq x < - \frac{1}{3}.\]
\[2)\log_{\frac{1}{3}}\left( \log_{4}\left( x^{2} - 5 \right) \right) > 0\]
\[\log_{\frac{1}{3}}\left( \log_{4}\left( x^{2} - 5 \right) \right) > \log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{0}\]
\[\log_{4}\left( x^{2} - 5 \right) < 1\]
\[\log_{4}\left( x^{2} - 5 \right) < \log_{4}4^{1}\]
\[x^{2} - 5 < 4\]
\[x^{2} < 9\]
\[- 3 < x < 3.\]
\[Имеет\ смысл\ при:\]
\[x^{2} - 5 > 0\]
\[x^{2} > 5\]
\[x < - \sqrt{5}\text{\ \ }и\ \ x > \sqrt{5}.\]
\[\log_{4}\left( x^{2} - 5 \right) > 0\]
\[\log_{4}\left( x^{2} - 5 \right) > \log_{4}4^{0}\]
\[x^{2} - 5 > 1\]
\[x^{2} > 6\]
\[x < - \sqrt{6}\text{\ \ }и\ \ x > \sqrt{6}.\]
\[Ответ:\ \]
\[- 3 < x < - \sqrt{6};\ \ \sqrt{6} < x < 3.\]
