ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 360

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 360

\[\boxed{\mathbf{360}\mathbf{.}}\]

\[1)\log_{8}\left( x^{2} - 4x + 3 \right) < 1\]

\[\log_{8}\left( x^{2} - 4x + 3 \right) < \log_{8}8\]

\[x^{2} - 4x + 3 < 8\]

\[x^{2} - 4x - 5 < 0\]

\[D = 4^{2} + 4 \bullet 5 = 16 + 20 = 36\]

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

\[x_{2} = \frac{4 + 6}{2} = 5.\]

\[(x + 1)(x - 5) < 0\]

\[- 1 < x < 5.\]

\[имеет\ смысл\ при:\]

\[x^{2} - 4x + 3 > 0\]

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\text{\ \ }x_{2} = \frac{4 + 2}{2} = 3.\]

\[(x - 1)(x - 3) > 0\]

\[x < 1;\text{\ \ }x > 3\]

\[Ответ:\ \ - 1 < x < 1;\ \ 3 < x < 5.\]

\[2)\log_{6}\left( x^{2} - 3x + 2 \right) \geq 1\]

\[\log_{6}\left( x^{2} - 3x + 2 \right) \geq \log_{6}6\]

\[x^{2} - 3x + 2 \geq 6\]

\[x^{2} - 3x - 4 \geq 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[(x + 1)(x - 4) \geq 0\]

\[x \leq - 1\ \ и\ \ x \geq 4.\]

\[имеет\ смысл\ при:\]

\[x^{2} - 3x + 2 > 0\]

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[x_{1} = \frac{3 - 1}{2} = 1;\text{\ \ }x_{2} = \frac{3 + 1}{2} = 2.\]

\[(x - 1)(x - 2) > 0\]

\[x < 1\ \ и\ \ x > 2.\]

\[Ответ:\ \ x \leq - 1;\ \ x \geq 4.\]

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

\[\log_{3}\left( x^{2} + 2x \right) > \log_{3}3\]

\[x^{2} + 2x > 3\]

\[x^{2} + 2x - 3 > 0\]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

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

\[x_{2} = \frac{- 2 + 4}{2} = 1.\]

\[(x + 3)(x - 1) > 0\]

\[x < - 3\ \ и\ \ x > 1.\]

\[имеет\ смысл\ при:\]

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

\[(x + 2)x > 0\]

\[x < - 2\ \ и\ \ x > 0.\]

\[Ответ:\ \ x < - 3;\ \ x > 1.\]

\[4)\log_{\frac{2}{3}}\left( x^{2} - 2,5x \right) < - 1\]

\[\log_{\frac{2}{3}}\left( x^{2} - 2,5x \right) < \log_{\frac{2}{3}}\left( \frac{2}{3} \right)^{- 1}\]

\[x^{2} - 2,5x > \left( \frac{2}{3} \right)^{- 1}\]

\[x^{2} - 2,5x > \frac{3}{2}\ \ \ \ \ | \bullet 2\]

\[2x^{2} - 5x > 3\]

\[2x^{2} - 5x - 3 > 0\]

\[D = 5^{2} + 4 \bullet 2 \bullet 3 =\]

\[= 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2 \bullet 2} = - \frac{2}{4} = - \frac{1}{2} = - 0,5;\]

\[x_{2} = \frac{5 + 7}{2 \bullet 2} = \frac{12}{4} = 3.\]

\[(x + 0,5)(x - 3) > 0\]

\[x < - 0,5\ \ и\ \ x > 3.\]

\[имеет\ смысл\ при:\]

\[x^{2} - 2,5x > 0\]

\[x(x - 2,5) > 0\]

\[x < 0\ \ и\ \ x > 2,5.\]

\[Ответ:\ \ x < - 0,5;\ \ x > 3.\]

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