Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 258

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\[1)\ {0,6}^{x} \bullet \left( \frac{25}{9} \right)^{x^{2} - 12} = \left( \frac{27}{125} \right)^{3}\ \]

\[\left( \frac{6}{10} \right)^{x} \bullet \left( \frac{5}{3} \right)^{2\left( x^{2} - 12 \right)} = \left( \frac{3}{5} \right)^{3 \bullet 3}\ \]

\[\left( \frac{3}{5} \right)^{x} \bullet \left( \frac{3}{5} \right)^{- 2\left( x^{2} - 12 \right)} = \left( \frac{3}{5} \right)^{9}\ \]

\[\left( \frac{3}{5} \right)^{x - 2\left( x^{2} - 12 \right)} = \left( \frac{3}{5} \right)^{9}\ \]

\[x - 2\left( x^{2} - 12 \right) = 9\ \]

\[x - 2x^{2} + 24 = 9\ \]

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

\[D = 1^{2} + 4 \bullet 2 \bullet 15 = 1 + 120 =\]

\[= 121\]

\[x_{1} = \frac{1 - 11}{2 \bullet 2} = - \frac{10}{4} = - 2,5;\ \]

\[x_{2} = \frac{1 + 11}{2 \bullet 2} = \frac{12}{4} = 3.\ \]

\[Ответ:\ \ x_{1} = - 2,5;\ \ \ x_{2} = 3.\ \ \]

\[2)\ 16\sqrt{{0,25}^{5 - \frac{x}{4}}} = 2^{\sqrt{x + 1}}\ \]

\[2^{4} \bullet \left( \frac{1}{4} \right)^{\frac{1}{2} \bullet \left( 5 - \frac{x}{4} \right)} = 2^{\sqrt{x + 1}}\ \]

\[2^{4} \bullet \left( 2^{- 2} \right)^{\frac{5}{2} - \frac{x}{8}} = 2^{\sqrt{x + 1}}\ \]

\[2^{4} \bullet 2^{\frac{x}{4} - 5} = 2^{\sqrt{x + 1}}\ \]

\[2^{4 + \frac{x}{4} - 5} = 2^{\sqrt{x + 1}}\ \]

\[4 + \frac{x}{4} - 5 = \sqrt{x + 1}\ \]

\[\frac{x}{4} - 1 = \sqrt{x + 1}\ \]

\[x - 4 = 4\sqrt{x + 1}\ \]

\[x^{2} - 8x + 16 = 16(x + 1)\ \]

\[x^{2} - 8x + 16 = 16x + 16\ \]

\[x^{2} - 24x = 0\ \]

\[x(x - 24) = 0\ \]

\[x_{1} = 0;\ \text{\ \ }x_{2} = 24.\ \]

\[Выражение\ имеет\ смысл:\]

\[x + 1 \geq 0\ \]

\[x \geq - 1.\ \]

\[Уравнение\ имеет\ решения:\]

\[x - 4 \geq 0\]

\[x \geq 4.\ \]

\[Ответ:\ \ x = 24.\]