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МатематикаРешебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 45
Задание 45
\[\boxed{\mathbf{45.}}\]
\[\textbf{а)}\ 2 + \log_{\sqrt{x^{2} - 2x - 3}}\left( \frac{x + 4}{x + 1} \right) >\]
\[> \log_{x^{2} - 2x - 3}\left( x^{2} - 2x - 2 \right)^{2}\]
\[x^{2} - 2x - 3 > 0\]
\[D_{1} = 1 + 3 = 4\]
\[x_{1} = 1 + 2 = 3;\]
\[x_{2} = 1 - 2 = - 1;\]
\[(x + 1)(x - 3) > 0\]
\[x < - 1;\ \ x > 3.\]
\[x^{2} - 2x - 3 \neq 1\]
\[x^{2} - 2x - 4 \neq 1\]
\[D_{1} = 1 + 5 = 5\]
\[x_{1} \neq 1 + \sqrt{5};\]
\[x_{2} \neq 1 - \sqrt{5}.\]
\[x^{2} - 2x - 2 \neq 0\]
\[D_{1} = 1 + 2 = 3\]
\[x_{1} \neq 1 + \sqrt{3};\]
\[x_{2} \neq 1 - \sqrt{3}.\]
\[\frac{x + 4}{x + 1} > 0\]
\[x < - 4;\ \ x > - 1.\]
\[2 + \log_{\sqrt{x^{2} - 2x - 3}}\left( \frac{x + 4}{x + 1} \right) =\]
\[= \log_{\sqrt{x^{2} - 2x - 3}}(x - 3)(x + 4) =\]
\[= \frac{\lg{(x - 3)(x - 4)}}{\lg\sqrt{x^{2} - 2x - 3}};\]
\[\log_{x^{2} - 2x - 3}\left( x^{2} - 2x - 2 \right)^{2} =\]
\[= \log_{\sqrt{x^{2} - 2x - 3}}\left( x^{2} - 2x - 3 \right) =\]
\[= \frac{\lg{(x^{2} - 2x - 2)}}{\lg\sqrt{x^{2} - 2x - 3}};\]
\[\frac{\lg{(x - 3)(x - 4)}}{\lg\sqrt{x^{2} - 2x - 3}} > \frac{\lg{(x^{2} - 2x - 2)}}{\lg\sqrt{x^{2} - 2x - 3}}\]
\[\lg\left( x^{2} + x - 12 \right) > \lg\left( x^{2} - 2x - 2 \right)\]
\[( - \infty; - 4) \cup \left( 1 + \sqrt{5}; + \infty \right):\]
\[x^{2} + x - 12 > x^{2} - 2x - 2\]
\[3x > 10\]
\[x > 3\frac{1}{3}.\]
\[\left( 3;1 + \sqrt{5} \right):\]
\[x^{2} + x - 12 < x^{2} - 2x - 2\]
\[3x < 10\]
\[x < 3\frac{1}{3}.\]
\[Решение\ неравенства:\]
\[x \in \left( 3;1 + \sqrt{5} \right) \cup \left( 3\frac{1}{3}; + \infty \right).\]
\[Ответ:x \in \left( 3;1 + \sqrt{5} \right) \cup \left( 3\frac{1}{3}; + \infty \right).\]
\[- x^{2} + 13x - 36 > 0\]
\[x^{2} - 13x + 36 < 0\]
\[x_{1} + x_{2} = 13;\ \ x_{1} \cdot x_{2} = 36\]
\[x_{1} = 4;\ \ x_{2} = 9\]
\[(x - 4)(x - 9) < 0\]
\[4 < x < 9.\]
\[- x^{2} + 13x - 36 \neq 1\]
\[x^{2} - 13x + 37 \neq 0\]
\[D = 169 - 148 = 21\]
\[x_{1} \neq \frac{13 + \sqrt{21}}{2};\]
\[x_{2} \neq \frac{13 - \sqrt{21}}{2}.\]
\[\frac{4,1 - x}{x - 9} > 0\]
\[4,1 < x < 9.\]
\[x^{2} + 10x + 32,93 \neq 1\]
\[x^{2} + 10x + 31,93 \neq 0\]
\[D = 100 - 127,72 < 0\]
\[x = R.\]
\[На\ множестве\ M_{1}:\]
\[На\ множестве\ M_{2}:\]
\[На\ множестве\ M_{3}:\]
\[Решение\ неравенства:\]
\[Ответ:\ \]
