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МатематикаРешебник по алгебре и начала математического анализа 11 класс Алимов Задание 361
Задание 361
\[\boxed{\mathbf{361}\mathbf{.}}\]
\[1)\lg\left( x^{2} - 8x + 13 \right) > 0\]
\[\lg\left( x^{2} - 8x + 13 \right) > \lg 1\]
\[x^{2} - 8x + 13 > 1\]
\[x^{2} - 8x + 12 > 0\]
\[D = 8^{2} - 4 \bullet 12 = 64 - 48 = 16\]
\[x_{1} = \frac{8 - 4}{2} = 2;\text{\ \ }x_{2} = \frac{8 + 4}{2} = 6.\]
\[(x - 2)(x - 6) > 0\]
\[x < 2\ \ и\ \ x > 6.\]
\[имеет\ смысл\ при:\]
\[x^{2} - 8x + 13 > 0\]
\[D = 8^{2} - 4 \bullet 13 = 64 - 52 = 12\]
\[x = \frac{8 \pm \sqrt{12}}{2} = \frac{8 \pm 2\sqrt{3}}{2} =\]
\[= 4 \pm \sqrt{3}.\]
\[x < 4 - \sqrt{3};\text{\ \ }x > 4 + \sqrt{3}.\]
\[Ответ:\ \ x < 2;\ \ x > 6.\]
\[2)\log_{\frac{1}{5}}\left( x^{2} - 5x + 7 \right) < 0\]
\[\log_{\frac{1}{5}}\left( x^{2} - 5x + 7 \right) < \log_{\frac{1}{5}}1\]
\[x^{2} - 5x + 7 > 1\]
\[x^{2} - 5x + 6 > 0\]
\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]
\[x_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }x_{2} = \frac{5 + 1}{2} = 3.\]
\[(x - 2)(x - 3) > 0\]
\[x < 2\ \ и\ \ x > 3.\]
\[\ имеет\ смысл\ при:\]
\[x^{2} - 5x + 7 > 0\]
\[D = 5^{2} - 4 \bullet 7 = 25 - 28 =\]
\[= - 3 < 0\]
\[a = 1 > 0 \Longrightarrow x - любое\ число.\]
\[Ответ:\ \ x < 2;\ \ x > 3.\]
\[3)\log_{2}\left( x^{2} + 2x \right) < 3\]
\[\log_{2}\left( x^{2} + 2x \right) < \log_{2}2^{3}\]
\[x^{2} + 2x < 2^{3}\]
\[x^{2} + 2x < 8\]
\[x^{2} + 2x - 8 < 0\]
\[D = 2^{2} + 4 \bullet 8 = 4 + 32 = 36\]
\[x_{1} = \frac{- 2 - 6}{2} = - 4;\text{\ \ }\]
\[x_{2} = \frac{- 2 + 6}{2} = 2.\]
\[(x + 4)(x - 2) < 0\]
\[- 4 < x < 2.\]
\[имеет\ смысл\ при:\]
\[x^{2} + 2x > 0\]
\[(x + 2) \bullet x > 0\]
\[x < - 2;\text{\ \ }x > 0.\]
\[Ответ:\ \ - 4 < x < - 2;\ \ \]
\[0 < x < 2.\]
\[4)\log_{\frac{1}{2}}\left( x^{2} - 5x - 6 \right) \geq - 3\]
\[\log_{\frac{1}{2}}\left( x^{2} - 5x - 6 \right) \geq \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{- 3}\]
\[x^{2} - 5x - 6 \leq \left( \frac{1}{2} \right)^{- 3}\]
\[x^{2} - 5x - 6 \leq 8\]
\[x^{2} - 5x - 14 \leq 0\]
\[D = 5^{2} + 4 \bullet 14 = 25 + 56 = 81\]
\[x_{1} = \frac{5 - 9}{2} = - 2;\text{\ \ }\]
\[x_{2} = \frac{5 + 9}{2} = 7.\]
\[(x + 2)(x - 7) \leq 0\]
\[- 2 \leq x \leq 7\]
\[имеет\ смысл\ при:\]
\[x^{2} - 5x - 6 > 0\]
\[D = 5^{2} + 4 \bullet 6 = 25 + 24 = 49\]
\[x_{1} = \frac{5 - 7}{2} = - 1;\text{\ \ }\]
\[x_{2} = \frac{5 + 7}{2} = 6.\]
\[(x + 1)(x - 6) > 0\]
\[x < - 1;\text{\ \ }x > 6.\]
\[Ответ:\ \ - 2 \leq x < - 1;\ \ \]
\[6 < x \leq 7.\]
