Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 401

\[\boxed{\mathbf{401.}}\]

\[1)\ x^{\lg 9} + 9^{\lg x} = 6\]

\[x^{\frac{\log_{x}9}{\log_{x}10}} + 9^{\lg x} = 6\]

\[\left( x^{\log_{x}9} \right)^{\frac{1}{\log_{x}10}} + 9^{\lg x} = 6\]

\[9^{\frac{1}{\log_{x}10}} + 9^{\lg x} = 6\]

\[9^{\frac{\log_{x}x}{\log_{x}10}} + 9^{\lg x} = 6\]

\[9^{\lg x} + 9^{\lg x} = 6\]

\[2 \bullet 9^{\lg x} = 6\]

\[9^{\lg x} = 3\]

\[\log_{9}9^{\lg x} = \log_{9}3\]

\[\lg x = \log_{9}9^{\frac{1}{2}}\]

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

\[\lg x = \lg 10^{\frac{1}{2}}\]

\[x = 10^{\frac{1}{2}} = \sqrt{10}\]

\[Ответ:\ \ x = \sqrt{10}.\]

\[2)\ x^{3\lg^{3}x - \frac{2}{3}\lg x} = 100\sqrt[3]{10}\]

\[\lg x^{3\lg^{3}x - \frac{2}{3}\lg x} = \lg\left( 100\sqrt[3]{10} \right)\]

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

\[= \lg\left( 10^{2} \bullet 10^{\frac{1}{3}} \right)\]

\[3\lg^{4}x - \frac{2}{3}\lg^{2}x = \lg 10^{\frac{7}{3}}\]

\[3\lg^{4}x - \frac{2}{3}\lg^{2}x = \frac{7}{3}\]

\[Пусть\ y = \lg^{2}x:\]

\[3y^{2} - \frac{2}{3}y = \frac{7}{3}\ \ \ \ \ | \bullet 3\]

\[9y^{2} - 2y - 7 = 0\]

\[D = 2^{2} + 4 \bullet 7 \bullet 9 =\]

\[= 4 + 252 = 256\]

\[y_{1} = \frac{2 - 16}{2 \bullet 9} = - \frac{14}{18} = - \frac{7}{8};\]

\[y_{2} = \frac{2 + 16}{2 \bullet 9} = \frac{18}{18} = 1.\]

\[1)\ \lg^{2}x = - \frac{7}{8}\]

\[нет\ корней.\]

\[2)\ \lg^{2}x = 1\]

\[\lg x = \pm 1\]

\[\lg x = \lg 10^{\pm 1}\]

\[x_{1} = 10^{- 1} = 0,1\]

\[x_{2} = 10^{1}\]

\[x = 10.\]

\[Ответ:\ \ x_{1} = 0,1;\ \ x_{2} = 10.\]