Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 217

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

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

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

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

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

\[4x^{2} - x - 3 = 0\]

\[D = 1^{2} + 4 \bullet 4 \bullet 3 = 1 + 48 = 49\]

\[x_{1} = \frac{1 - 7}{2 \bullet 4} = - \frac{6}{8} = - 0,75;\ \]

\[x_{2} = \frac{1 + 7}{2 \bullet 4} = \frac{8}{8} = 1.\]

\[Ответ:\ \ x_{1} = - 0,75;\ \ \ x_{2} = 1.\]

\[2)\ 5^{0,1x} \bullet \left( \frac{1}{5} \right)^{- 0,06} = 5^{x^{2}}\]

\[5^{0,1x} \bullet 5^{0,06} = 5^{x^{2}}\]

\[5^{0,1x + 0,06} = 5^{x^{2}}\]

\[0,1x + 0,06 = x^{2}\]

\[10x + 6 = 100x^{2}\]

\[100x^{2} - 10x - 6 = 0\]

\[D = 10^{2} + 4 \bullet 100 \bullet 6 =\]

\[= 100 + 2400 = 2500\]

\[x_{1} = \frac{10 - 50}{2 \bullet 100} = - \frac{40}{200} = - 0,2;\ \]

\[x_{2} = \frac{10 + 50}{2 \bullet 100} = \frac{60}{200} = 0,3.\]

\[Ответ:\ \ x_{1} = - 0,2;\ \ \ x_{2} = 0,3.\]

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

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

\[\sqrt{1 - x} - 1 = 2x\]

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

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

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

\[x(4x + 5) = 0\]

\[x_{1} = 0;\ \text{\ \ }x_{2} = - \frac{5}{4} = - 1,25.\]

\[Выражение\ имеет\ смысл\ при:\]

\[1 - x \geq 0\]

\[x \leq 1.\]

\[Уравнение\ имеет\ решения\ при:\]

\[2x + 1 \geq 0\]

\[2x \geq - 1\]

\[x \geq - \frac{1}{2}.\]

\[Ответ:\ \ x = 0.\]

\[4)\ {0,7}^{\sqrt{x + 12}} \bullet {0,7}^{- 2} = {0,7}^{\sqrt{x}}\]

\[{0,7}^{\sqrt{x + 12} - 2} = {0,7}^{\sqrt{x}}\]

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

\[\sqrt{x + 12} = \sqrt{x} + 2\]

\[x + 12 = x + 4\sqrt{x} + 4\]

\[8 = 4\sqrt{x}\]

\[2 = \sqrt{x}\]

\[x = 4.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x + 12 \geq 0\]

\[x \geq - 12.\]

\[Ответ:\ \ x = 4.\]