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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 392
Задание 392
\[\boxed{\mathbf{392}\mathbf{.}}\]
\[1)\log_{3}\left( 2 - x^{2} \right) - \log_{3}( - x) = 0\]
\[\log_{3}\left( 2 - x^{2} \right) = \log_{3}( - x)\]
\[2 - x^{2} = - x\]
\[x^{2} - x - 2 = 0\]
\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]
\[x_{1} = \frac{1 - 3}{2} = - 1;\text{\ \ }\]
\[x_{2} = \frac{1 + 3}{2} = 2.\]
\[имеет\ смысл\ при:\]
\[1)\ 2 - x^{2} > 0\]
\[x^{2} < 2\]
\[- \sqrt{2} < x < \sqrt{2}.\]
\[2)\ - x > 0\ \]
\[x < 0.\]
\[Ответ:\ \ x = - 1.\]
\[2)\log_{5}\left( x^{2} - 12 \right) - \log_{5}( - x) = 0\]
\[\log_{5}\left( x^{2} - 12 \right) = \log_{5}( - x)\]
\[x^{2} - 12 = - x\]
\[x^{2} + x - 12 = 0\]
\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]
\[x_{1} = \frac{- 1 - 7}{2} = - 4;\text{\ \ }\]
\[x_{2} = \frac{- 1 + 7}{2} = 3.\]
\[имеет\ смысл\ при:\]
\[1)\ x^{2} - 12 > 0\]
\[x^{2} > 12\]
\[x < - 2\sqrt{3}\text{\ \ }и\ \ x > 2\sqrt{3}.\]
\[2)\ - x > 0\ \]
\[x < 0.\]
\[Ответ:\ \ x = - 4.\]
\[3)\log_{2}\sqrt{x - 3} + \log_{2}\sqrt{3x - 7} =\]
\[= 2\]
\[\log_{2}\sqrt{(x - 3)(3x - 7)} = \log_{2}2^{2}\]
\[\sqrt{(x - 3)(3x - 7)} = 2^{2}\]
\[\sqrt{3x^{2} - 7x - 9x + 21} = 4\]
\[3x^{2} - 16x + 21 = 16\]
\[3x^{2} - 16x + 5 = 0\]
\[D = 16^{2} - 4 \bullet 3 \bullet 5 =\]
\[= 256 - 60 = 196\]
\[x_{1} = \frac{16 - 14}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]
\[x_{2} = \frac{16 + 14}{2 \bullet 3} = \frac{30}{6} = 5.\]
\[имеет\ смысл\ при:\]
\[x - 3 > 0 \Longrightarrow \ x > 3;\]
\[3x - 7 > 0 \Longrightarrow x > 2\frac{1}{3}.\]
\[Ответ:\ \ x = 5.\]
\[4)\lg(x + 6) - \lg\sqrt{2x - 3} = \lg 4\]
\[\lg\frac{x + 6}{\sqrt{2x - 3}} = \lg 4\]
\[\frac{x + 6}{\sqrt{2x - 3}} = 4\]
\[x + 6 = 4\sqrt{2x - 3}\]
\[x^{2} + 12x + 36 = 16(2x - 3)\]
\[x^{2} + 12x + 36 = 32x - 48\]
\[x^{2} - 20x + 84 = 0\]
\[D = 20^{2} - 4 \bullet 84 =\]
\[= 400 - 336 = 64\]
\[x_{1} = \frac{20 - 8}{2} = 6;\ \]
\[x_{2} = \frac{20 + 8}{2} = 14.\]
\[имеет\ смысл\ при:\]
\[x + 6 > 0 \Longrightarrow x > - 6;\]
\[2x - 3 > 0 \Longrightarrow x > 1,5.\]
\[Ответ:\ \ x_{1} = 6;\ \ x_{2} = 14.\]
