Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 742

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\[1)\ {8,4}^{\frac{x - 3}{x^{2} + 1}} < 1\ \]

\[{8,4}^{\frac{x - 3}{x^{2} + 1}} < {8,4}^{0}\ \]

\[\frac{x - 3}{x^{2} + 1} < 0\ \]

\[x - 3 < 0\]

\[x < 3\ \]

\[Ответ:\ \ x < 3.\]

\[2)\ 2^{x^{2}} \bullet 5^{x^{2}} < 10^{- 3} \bullet \left( 10^{3 - x} \right)^{2}\ \]

\[(2 \bullet 5)^{x^{2}} < 10^{- 3} \bullet 10^{6 - 2x}\ \]

\[10^{x^{2}} < 10^{- 3 + 6 - 2x}\ \]

\[x^{2} < - 3 + 6 - 2x\ \]

\[x^{2} + 2x - 3 < 0\ \]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[x_{1} = \frac{- 2 - 4}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{- 2 + 4}{2} = 1.\ \]

\[(x + 3)(x - 1) < 0\ \]

\[- 3 < x < 1\ \]

\[Ответ:\ \ - 3 < x < 1.\]

\[3)\ \frac{4^{x} - 2^{x + 1} + 8}{2^{1 - x}} < 8^{x}\ \]

\[4^{x} - 2^{x + 1} + 8 < 8^{x} \bullet 2^{1 - x}\ \]

\[2^{2x} - 2 \bullet 2^{x} + 8 < 2^{3x} \bullet 2 \bullet 2^{- x}\ \]

\[2^{2x} - 2 \bullet 2^{x} + 8 < 2 \bullet 2^{2x}\ \]

\[2^{2x} + 2 \bullet 2^{x} - 8 > 0\ \]

\[Пусть\ y = 2^{x}:\]

\[y^{2} + 2y - 8 > 0\ \]

\[D = 2^{2} + 4 \bullet 8 = 4 + 32 = 36\]

\[y_{1} = \frac{- 2 - 6}{2} = - 4;\]

\[\text{\ \ }y_{2} = \frac{- 2 + 6}{2} = 2.\ \]

\[(y + 4)(y - 2) > 0\ \]

\[y < - 4\ \ и\ \ y > 2.\ \]

\[1)\ 2^{x} < - 4\]

\[нет\ корней.\ \]

\[2)\ 2^{x} > 2\ \]

\[2^{x} > 2^{1}\ \]

\[x > 1.\ \]

\[Ответ:\ \ x > 1.\]

\[4)\ \frac{1}{3^{x} + 5} \leq \frac{1}{3^{x + 1} - 1}\ \]

\[3^{x} + 5 \geq 3^{x + 1} - 1\ \]

\[3^{x + 1} - 3^{x} \leq 6\ \]

\[3^{x} \bullet \left( 3^{1} - 1 \right) \leq 6\ \]

\[3^{x} \bullet 2 \leq 6\ \]

\[3^{x} \leq 3\ \]

\[3^{x} \leq 3^{1}\ \]

\[x \leq 1.\ \]

\[Неравенство\ имеет\ \]

\[решения\ при:\]

\[3^{x + 1} - 1 \geq 0\ \]

\[3^{x + 1} \geq 1\ \]

\[3^{x + 1} \geq 3^{0}\ \]

\[x + 1 \geq 0\]

\[x \geq - 1.\ \]

\[Ответ:\ \ - 1 \leq x \leq 1.\]