ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 259

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\[1)\ 2 \bullet 3^{3x - 1} + 27^{x - \frac{2}{3}} =\]

\[= 9^{x - 1} + 2 \bullet 3^{2x - 1}\ \]

\[2 \bullet 3^{3x} \bullet \frac{1}{3} + \left( 3^{3} \right)^{x - \frac{2}{3}} =\]

\[= \left( 3^{2} \right)^{x - 1} + 2 \bullet 3^{2x} \bullet \frac{1}{3}\ \]

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

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

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

\[3x = 2x\ \]

\[x = 0\ \]

\[Ответ:\ \ x = 0.\]

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

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

\[2^{\sqrt{x}} \bullet \left( 2^{2} - 2^{1} - 2^{- 1} \right) = 12\ \]

\[2^{\sqrt{x}} \bullet \left( 4 - 2 - \frac{1}{2} \right) = 12\ \]

\[2^{\sqrt{x}} \bullet \left( 2 - \frac{1}{2} \right) = 12\ \]

\[2^{\sqrt{x}} \bullet \frac{3}{2} = 12\ \]

\[2^{\sqrt{x}} = 8\ \]

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

\[\sqrt{x} = 3\ \]

\[x = 9\ \]

\[Ответ:\ \ x = 9.\]

\[22 \bullet 3^{2x} - 81 \bullet 3^{x} + 27 \bullet 3^{x} = 36\ \]

\[22 \bullet 3^{2x} - 54 \bullet 3^{x} - 36 = 0\ \]

\[Пусть\ y = 3^{x}:\]

\[22y^{2} - 54y - 36 = 0\ \]

\[D = 54^{2} + 4 \bullet 22 \bullet 36 =\]

\[= 2916 + 3168 = 6084\]

\[y_{1} = \frac{54 - 78}{2 \bullet 22} = - \frac{24}{44} = - \frac{6}{11};\ \]

\[y_{2} = \frac{54 + 78}{2 \bullet 22} = \frac{132}{44} = 3.\ \]

\[1)\ 3^{x} = - \frac{6}{11}\]

\[нет\ корней.\ \]

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

\[x = 1.\ \]

\[Ответ:\ \ x = 1.\]

\[\frac{5}{4} \bullet 4^{x} - 4^{2x} + 4^{x} + 7 = 0\ \ \ \ \ | \bullet 4\ \]

\[5 \bullet 4^{x} - 4 \bullet 4^{2x} + 4 \bullet 4^{x} + 28 = 0\ \]

\[4 \bullet 4^{2x} - 9 \bullet 4^{x} - 28 = 0\ \]

\[Пусть\ y = 4^{x}:\]

\[4y^{2} - 9y - 28 = 0\ \]

\[D = 9^{2} + 4 \bullet 4 \bullet 28 =\]

\[= 81 + 448 = 529\]

\[y_{1} = \frac{9 - 23}{2 \bullet 4} = - \frac{14}{8} = - \frac{7}{4};\ \]

\[y_{2} = \frac{9 + 23}{2 \bullet 4} = \frac{32}{8} = 4.\ \]

\[1)\ 4^{x} = - \frac{7}{4}\]

\[нет\ корней.\]

\[2)\ 4^{x} = 4\ \]

\[x = 1.\ \]

\(Ответ:\ \ x = 1.\)