Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 396

\[\boxed{\mathbf{396}\mathbf{.}}\]

\[(x - 4)(x + 1) \leq \left( \sqrt{6} \right)^{2}\]

\[x^{2} + x - 4x - 4 \leq 6\]

\[x^{2} - 3x - 10 \leq 0\]

\[D = 3^{2} + 4 \bullet 10 = 9 + 40 = 49\]

\[x_{1} = \frac{3 - 7}{2} = - 2;\text{\ \ }\]

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

\[(x + 2)(x - 5) \leq 0\]

\[- 2 \leq x \leq 5.\]

\[имеет\ смысл\ при:\]

\[x - 4 > 0 \Longrightarrow x > 4;\]

\[x + 1 > 0 \Longrightarrow x > - 1.\]

\[Ответ:\ \ 4 < x \leq 5.\]

\[(x - 5)(x + 12) \leq \left( 3\sqrt{2} \right)^{2}\]

\[x^{2} + 12x - 5x - 60 \leq 9 \bullet 2\]

\[x^{2} + 7x - 78 \leq 0\]

\[D = 7^{2} + 4 \bullet 78 = 49 + 312 =\]

\[= 361\]

\[x_{1} = \frac{- 7 - 19}{2} = - 13;\ \ \]

\[x_{2} = \frac{- 7 + 19}{2} = 6.\]

\[(x + 13)(x - 6) \leq 0\]

\[- 13 \leq x \leq 6.\]

\[имеет\ смысл\ при:\]

\[x - 5 > 0 \Longrightarrow x > 5;\]

\[x + 12 > 0 \Longrightarrow x > - 12.\]

\[Ответ:\ \ 5 < x \leq 6.\]

\[\log_{3}\frac{8x^{2} + x}{x^{2} \bullet x} > \log_{3}3^{2}\]

\[\log_{3}\frac{8x + 1}{x^{2}} > \log_{3}9\]

\[\frac{8x + 1}{x^{2}} > 9\]

\[8x + 1 > 9x^{2}\]

\[9x^{2} - 8x - 1 < 0\]

\[D = 8^{2} + 4 \bullet 9 = 64 + 36 = 100\]

\[x_{1} = \frac{8 - 10}{2 \bullet 9} = - \frac{2}{18} = - \frac{1}{9};\]

\[x_{2} = \frac{8 + 10}{2 \bullet 9} = \frac{18}{18} = 1.\]

\[\left( x + \frac{1}{9} \right)(x - 1) < 0\]

\[- \frac{1}{9} < x < 1.\]

\[имеет\ смысл\ при:\]

\[8x^{2} + x > 0\]

\[(8x + 1)x > 0\]

\[x < - \frac{1}{8}\text{\ \ }и\ \ x > 0.\]

\[Ответ:\ \ 0 < x < 1.\]

\[4)\log_{2}x + \log_{2}(x - 3) > \log_{2}4\]

\[\log_{2}\left( x(x - 3) \right) > \log_{2}4\]

\[x(x - 3) > 4\]

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

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

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

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

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

\[x < - 1\ \ и\ \ x > 4.\]

\[имеет\ смысл\ при:\]

\[x > 0;\]

\[x - 3 > 0 \Longrightarrow x > 3.\]

\[Ответ:\ \ x > 4.\]

\[\log_{\frac{1}{5}}\frac{x - 10}{x + 2} \geq \log_{\frac{1}{5}}\left( \frac{1}{5} \right)^{- 1}\]

\[\frac{x - 10}{x + 2} \leq \left( \frac{1}{5} \right)^{- 1}\]

\[(x - 10)(x + 2) \leq 5(x + 2)^{2}\]

\[x^{2} - 8x - 20 \leq 5x^{2} + 20x + 20\]

\[4x^{2} + 28x + 40 \geq 0\]

\[x^{2} + 7x + 10 \geq 0\]

\[D = 7^{2} - 4 \bullet 10 = 49 - 40 = 9\]

\[x_{1} = \frac{- 7 - 3}{2} = - 5;\text{\ \ }\]

\[x_{2} = \frac{- 7 + 3}{2} = - 2.\]

\[(x + 5)(x + 2) \geq 0\]

\[x \leq - 5\ \ и\ \ x \geq - 2.\]

\[имеет\ смысл\ при:\]

\[x - 10 > 0 \Longrightarrow \ x > 10;\]

\[x + 2 > 0 \Longrightarrow x > - 2.\]

\[Ответ:\ \ x > 10.\]

\[(x + 10)(x + 4) < \left( \frac{1}{\sqrt{7}} \right)^{- 2}\]

\[x^{2} + 4x + 10x + 40 < \left( \sqrt{7} \right)^{2}\]

\[x^{2} + 14x + 40 < 7\]

\[x^{2} + 14x + 33 < 0\]

\[D = 14^{2} - 4 \bullet 33 = 196 - 132 =\]

\[= 64\]

\[x_{1} = \frac{- 14 - 8}{2} = - 11;\text{\ \ }\]

\[x_{2} = \frac{- 14 + 8}{2} = - 3.\]

\[(x + 11)(x + 3) < 0\]

\[- 11 < x < - 3.\]

\[имеет\ смысл\ при:\]

\[x + 10 > 0 \Longrightarrow x > - 10;\]

\[x + 4 > 0 \Longrightarrow x > - 4.\]

\[Ответ:\ \ - 4 < x < - 3.\]