Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 686

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[- x = - 3;\]

\[x = 3;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x - 2 \neq 0 \Longrightarrow x \neq 2.\]

\[Ответ:\ \ x = 3.\]

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

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

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

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

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

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

\[Выражение\ имеет\ смысл\ при:\]

\[x \neq 0;\]

\[x + 1 \neq 0\ \Longrightarrow x \neq - 1.\]

\[Ответ:\ \ x = - \frac{1}{3}.\]