Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 685

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\[1)\ 3^{x} + 3^{3 - x} - 12 = 0\]

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

\[3^{x} = t > 0:\]

\[t + \frac{27}{t} - 12 = 0\ \ \ \ \ | \cdot t\]

\[t^{2} - 12t + 27 = 0\]

\[D_{1} = 36 - 27 = 9\]

\[t_{1} = 6 + 3 = 9;\ \ t_{2} = 6 - 3 = 3.\]

\[t = 9:\]

\[3^{x} = 9\]

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

\[x = 2.\]

\[t = 3:\]

\[3^{x} = 3\]

\[x = 1.\]

\[Ответ:x = 1;\ \ x = 2.\]

\[2)\ 2^{x + 2} - 2^{2 - x} = 15\]

\[2^{x} \cdot 2^{2} - 2^{2} \cdot 2^{- x} - 15 = 0\]

\[t = 2^{x} > 0:\]

\[4t - \frac{4}{t} - 15 = 0\ \ \ \ | \cdot t\]

\[4t^{2} - 15t - 4 = 0\]

\[D = 225 + 64 = 289 = 17^{2}\]

\[t_{1} = \frac{15 - 17}{8} =\]

\[= - \frac{1}{4} < 0\ (не\ подходит);\]

\[t_{2} = \frac{15 + 17}{8} = 4.\]

\[t = 4:\]

\[2^{x} = 4\]

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

\[x = 2.\]

\[Ответ:x = 2.\]