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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 360
Задание 360
\[\boxed{\mathbf{360}\mathbf{.}}\]
\[1)\log_{8}\left( x^{2} - 4x + 3 \right) < 1\]
\[\log_{8}\left( x^{2} - 4x + 3 \right) < \log_{8}8\]
\[x^{2} - 4x + 3 < 8\]
\[x^{2} - 4x - 5 < 0\]
\[D = 4^{2} + 4 \bullet 5 = 16 + 20 = 36\]
\[x_{1} = \frac{4 - 6}{2} = - 1;\ \]
\[x_{2} = \frac{4 + 6}{2} = 5.\]
\[(x + 1)(x - 5) < 0\]
\[- 1 < x < 5.\]
\[имеет\ смысл\ при:\]
\[x^{2} - 4x + 3 > 0\]
\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]
\[x_{1} = \frac{4 - 2}{2} = 1;\text{\ \ }x_{2} = \frac{4 + 2}{2} = 3.\]
\[(x - 1)(x - 3) > 0\]
\[x < 1;\text{\ \ }x > 3\]
\[Ответ:\ \ - 1 < x < 1;\ \ 3 < x < 5.\]
\[2)\log_{6}\left( x^{2} - 3x + 2 \right) \geq 1\]
\[\log_{6}\left( x^{2} - 3x + 2 \right) \geq \log_{6}6\]
\[x^{2} - 3x + 2 \geq 6\]
\[x^{2} - 3x - 4 \geq 0\]
\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]
\[x_{1} = \frac{3 - 5}{2} = - 1;\text{\ \ }\]
\[x_{2} = \frac{3 + 5}{2} = 4.\]
\[(x + 1)(x - 4) \geq 0\]
\[x \leq - 1\ \ и\ \ x \geq 4.\]
\[имеет\ смысл\ при:\]
\[x^{2} - 3x + 2 > 0\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[x_{1} = \frac{3 - 1}{2} = 1;\text{\ \ }x_{2} = \frac{3 + 1}{2} = 2.\]
\[(x - 1)(x - 2) > 0\]
\[x < 1\ \ и\ \ x > 2.\]
\[Ответ:\ \ x \leq - 1;\ \ x \geq 4.\]
\[3)\log_{3}\left( x^{2} + 2x \right) > 1\]
\[\log_{3}\left( x^{2} + 2x \right) > \log_{3}3\]
\[x^{2} + 2x > 3\]
\[x^{2} + 2x - 3 > 0\]
\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]
\[x_{1} = \frac{- 2 - 4}{2} = - 3;\text{\ \ }\]
\[x_{2} = \frac{- 2 + 4}{2} = 1.\]
\[(x + 3)(x - 1) > 0\]
\[x < - 3\ \ и\ \ x > 1.\]
\[имеет\ смысл\ при:\]
\[x^{2} + 2x > 0\]
\[(x + 2)x > 0\]
\[x < - 2\ \ и\ \ x > 0.\]
\[Ответ:\ \ x < - 3;\ \ x > 1.\]
\[4)\log_{\frac{2}{3}}\left( x^{2} - 2,5x \right) < - 1\]
\[\log_{\frac{2}{3}}\left( x^{2} - 2,5x \right) < \log_{\frac{2}{3}}\left( \frac{2}{3} \right)^{- 1}\]
\[x^{2} - 2,5x > \left( \frac{2}{3} \right)^{- 1}\]
\[x^{2} - 2,5x > \frac{3}{2}\ \ \ \ \ | \bullet 2\]
\[2x^{2} - 5x > 3\]
\[2x^{2} - 5x - 3 > 0\]
\[D = 5^{2} + 4 \bullet 2 \bullet 3 =\]
\[= 25 + 24 = 49\]
\[x_{1} = \frac{5 - 7}{2 \bullet 2} = - \frac{2}{4} = - \frac{1}{2} = - 0,5;\]
\[x_{2} = \frac{5 + 7}{2 \bullet 2} = \frac{12}{4} = 3.\]
\[(x + 0,5)(x - 3) > 0\]
\[x < - 0,5\ \ и\ \ x > 3.\]
\[имеет\ смысл\ при:\]
\[x^{2} - 2,5x > 0\]
\[x(x - 2,5) > 0\]
\[x < 0\ \ и\ \ x > 2,5.\]
\[Ответ:\ \ x < - 0,5;\ \ x > 3.\]
