Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 214

\[\boxed{\mathbf{214}\mathbf{.}}\]

\[1)\ 3^{x^{2} + x - 12} = 1\]

\[3^{x^{2} + x - 12} = 3^{0}\]

\[x^{2} + x - 12 = 0\]

\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]

\[x_{1} = \frac{- 1 - 7}{2} = - 4;\ \ \text{\ \ }\]

\[x_{2} = \frac{- 1 + 7}{2} = 3\]

\[Ответ:\ \ x_{1} = - 4;\ \ \ x_{2} = 3.\]

\[2)\ 2^{x^{2} - 7x + 10} = 1\]

\[2^{x^{2} - 7x + 10} = 2^{0}\]

\[x^{2} - 7x + 10 = 0\]

\[D = 7^{2} - 4 \bullet 10 = 49 - 40 = 9\]

\[x_{1} = \frac{7 - 3}{2} = 2;\ \text{\ \ }x_{2} = \frac{7 + 3}{2} = 5\]

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

\[3)\ 2^{\frac{x - 1}{x - 2}} = 4\]

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

\[\frac{x - 1}{x - 2} = 2;\ \ \ \ x \neq 2\]

\[x - 1 = 2(x - 2)\]

\[x - 1 = 2x - 4\]

\[- x = - 3\]

\[x = 3.\]

\[Ответ:\ \ x = 3.\]

\[4)\ {0,5}^{\frac{1}{x}} = 4^{\frac{1}{x + 1}}\]

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

\[2^{- \frac{1}{x}} = 2^{\frac{2}{x + 1}}\]

\[- \frac{1}{x} = \frac{2}{x + 1};\ \ \ \ \ \ \ x \neq 0;\ \ \ x \neq - 1\]

\[x + 1 = - 2x\ \]

\[3x = - 1\]

\[x = - \frac{1}{3}.\]

\[Ответ:\ \ x = - \frac{1}{3}.\]