ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 853

\[1)\ 9 \bullet 4^{\frac{1}{x}} + 5 \bullet 6^{\frac{1}{x}} = 4 \bullet 9^{\frac{1}{x}}\]

\[9 \bullet 2^{\frac{2}{x}} + 5 \bullet 2^{\frac{1}{x}} \bullet 3^{\frac{1}{x}} - 4 \bullet 3^{\frac{2}{x}} = 0\ \ \ \ \ |\ :3^{\frac{2}{x}}\]

\[9 \bullet \left( \frac{2}{3} \right)^{\frac{2}{x}} + 5 \bullet \left( \frac{2}{3} \right)^{\frac{1}{x}} - 4 = 0\]

\[D = 25 + 144 = 169\]

\[\left( \frac{2}{3} \right)_{1}^{\frac{1}{x}} = \frac{- 5 - 13}{2 \bullet 9} < 0;\text{\ \ }\]

\[\left( \frac{2}{3} \right)_{2}^{\frac{1}{x}} = \frac{- 5 + 13}{2 \bullet 9} = \frac{4}{9};\]

\[\left( \frac{2}{3} \right)^{\frac{1}{x}} = \frac{4}{9} = \left( \frac{2}{3} \right)^{2}\]

\[\frac{1}{x} = 2\]

\[x = \frac{1}{2} = 0,5.\]

\[Ответ:\ \ 0,5.\]

\[2)\log_{2}\left( x^{2} - 3 \right) - \log_{2}(6x - 10) + 1 = 0\]

\[\log_{2}\left( x^{2} - 3 \right) - \log_{2}(6x - 10) = - 1\]

\[\log_{2}\frac{x^{2} - 3}{6x - 10} = \log_{2}\frac{1}{2}\]

\[\frac{x^{2} - 3}{6x - 10} = \frac{1}{2}\]

\[2\left( x^{2} - 3 \right) = 6x - 10\]

\[2x^{2} - 6 = 6x - 10\]

\[2x^{2} - 6x + 4 = 0\]

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

\[D = 9 - 8 = 1\]

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

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

\[Область\ определения:\]

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

\[x < - \sqrt{3};\ \ \ x > \sqrt{3}.\]

\[6x - 10 > 0\text{\ \ \ }\]

\[x > \frac{5}{3}.\]

\[Ответ:\ \ 2.\]

\[3)\ 2\log_{2}x - 2\log_{2}\frac{1}{\sqrt{2}} = 3\sqrt{\log_{2}x}\]

\[2\log_{2}x - \log_{2}\frac{1}{2} = 3\sqrt{\log_{2}x}\]

\[2\log_{2}x - 3\sqrt{\log_{2}x} + 1 = 0\]

\[D = 9 - 8 = 1\]

\[\sqrt{\log_{2}x_{1}} = \frac{3 - 1}{2 \bullet 2} = \frac{1}{2};\]

\[\sqrt{\log_{2}x_{2}} = \frac{3 + 1}{2 \bullet 2} = 1.\]

\[\log_{2}x_{1} = \frac{1}{2^{2}} = \frac{1}{4};\]

\[\log_{2}x_{2} = 1^{2} = 1;\]

\[x_{1} = 2^{\frac{1}{4}} = \sqrt[4]{2}\ ;\]

\[x_{2} = 2^{1} = 2.\]

\[Ответ:\ \ \sqrt[4]{2};\ 2.\]

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

\[\frac{\lg\left( 2x^{2} - 3x - 4 \right)}{\lg x} = 2\]

\[\lg\left( 2x^{2} - 3x - 4 \right) = 2\lg x\]

\[\lg\left( 2x^{2} - 3x - 4 \right) = \lg x^{2}\]

\[2x^{2} - 3x - 4 = x^{2}\]

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

\[D = 9 + 16 = 25\]

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

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

\[Область\ определения:\]

\[x > 0;\ \ \ x \neq 1.\]

\[Ответ:\ \ 4.\]