Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 889

\[\boxed{\mathbf{889}.}\]

\[1)\lg\left( x^{2} - 2x \right) = \lg 30 - 1\]

\[\lg\left( x^{2} - 2x \right) = \lg 30 - \lg 10\]

\[\lg\left( x^{2} - 2x \right) = \lg\frac{30}{10}\]

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

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

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

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

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

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

\[имеет\ смысл\ при:\]

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

\[x(x - 2) > 0\]

\[x < 0;\text{\ \ }x > 2\]

\[Ответ:\ \ x_{1} = - 1;\ \ x_{2} = 3.\]

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

\[= \log_{3}6 - \log_{3}2\]

\[\log_{3}\left( 2x^{2} + x \right) = \log_{3}\frac{6}{2}\]

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

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

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

\[D = 1^{2} + 4 \bullet 2 \bullet 3 = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2 \bullet 2} = - \frac{6}{4} = - 1,5;\]

\[x_{2} = \frac{- 1 + 5}{2 \bullet 2} = \frac{4}{4} = 1.\]

\[имеет\ смысл\ при:\]

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

\[(2x + 1)x > 0\]

\[x < - 0,5;\text{\ \ }x > 0.\]

\[Ответ:\ \ x_{1} = - 1,5;\ \ x_{2} = 1.\]

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

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

\[Пусть\ y = \lg x:\]

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

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

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

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

\[1)\ \lg x = - 1\]

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

\[x = 10^{- 1}\]

\[x = 0,1.\]

\[2)\ \lg x = 4\]

\[\lg x = \lg 10^{4}\]

\[x = 10^{4} = 10\ 000.\]

\[Ответ:\ \ x_{1} = 0,1;\ \ x_{2} = 10\ 000.\]

\[4)\log_{2}^{2}x - 5\log_{2}x + 6 = 0\]

\[Пусть\ y = \log_{2}x:\]

\[y^{2} - 5y + 6 = 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }y_{2} = \frac{5 + 1}{2} = 3.\]

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

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

\[x = 2^{2}\]

\[x = 4.\]

\[2)\ \log_{2}x = 3\]

\[\log_{2}x = \log_{2}2^{3}\]

\[x = 2^{3}\]

\[x = 8\]

\[Ответ:\ \ x_{1} = 4;\ \ x_{2} = 8.\]