ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1597

\[\boxed{\mathbf{1597}\mathbf{.}}\]

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

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

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

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

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

\[9y^{2} + 5y - 4 = 0\]

\[D = 25 + 144 = 169\]

\[y_{1} = \frac{- 5 - 13}{2 \bullet 9} = - \frac{18}{18} = - 1;\]

\[y_{2} = \frac{- 5 + 13}{2 \bullet 9} = \frac{8}{18} = \frac{4}{9}.\]

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

\[корней\ нет.\]

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

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

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

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

\[Ответ:\ \ x = \frac{1}{2}.\]

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

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

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

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

\[2x^{2} - 6 - 6x + 10 = 0\]

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

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

\[D = 9 - 8 = 1\]

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

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

\[Имеет\ смысл\ при:\]

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

\[x^{2} > 3\]

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

\[6x - 10 > 0\]

\[3x - 5 > 0\]

\[3x > 5\]

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

\[Ответ:\ \ x = 2.\]

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

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

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

\[y = \sqrt{\log_{2}x}:\]

\[2y^{2} - 3y + 1 = 0\]

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

\[y_{1} = \frac{3 - 1}{2 \bullet 2} = \frac{1}{2};\]

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

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

\[\log_{2}x = \frac{1}{4}\]

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

\[x = \sqrt[4]{2}.\]

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

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

\[\log_{2}x = \log_{2}2^{1}\]

\[x = 2.\]

\[Имеет\ смысл\ при:\]

\[x > 0.\]

\[\log_{2}x \geq 0\]

\[\log_{2}x \geq \log_{2}2^{0}\]

\[x \geq 1.\]

\[Ответ:\ \ x = \sqrt[4]{2};\ \ x = 2.\]

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

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

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

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

\[D = 9 + 16 = 25\]

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

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

\[Имеет\ смысл\ при:\]

\[x > 0\ \ и\ \ x \neq 1.\]

\[Проверка:\]

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

\[= \log_{4}(32 - 12 - 4) =\]

\[= \log_{4}16 = 2.\]

\[Ответ:\ \ x = 4.\]