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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 706
Задание 706
\[\boxed{\mathbf{706}.}\]
\[1)\ 2^{- x^{2} + 3x} < 4;\]
\[2^{- x^{2} + 3x} < 2^{2};\]
\[- x^{2} + 3x < 2;\]
\[x^{2} - 3x + 2 > 0;\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[x_{1} = \frac{3 - 1}{2} = 1\ \ и\ \]
\[\ x_{2} = \frac{3 + 1}{2} = 2;\]
\[(x - 1)(x - 2) > 0;\]
\[x < 1\ \ и\ \ x > 2;\]
\[Ответ:\ \ x < 1;\ \ x > 2.\]
\[2)\ \left( \frac{7}{9} \right)^{2x^{2} - 3x} \geq \frac{9}{7};\]
\[\left( \frac{7}{9} \right)^{2x^{2} - 3x} \geq \left( \frac{7}{9} \right)^{- 1};\]
\[2x^{2} - 3x \leq - 1;\]
\[2x^{2} - 3x + 1 \leq 0;\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[x_{1} = \frac{3 - 1}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2} = 0,5;\]
\[x_{2} = \frac{3 + 1}{2 \bullet 2} = \frac{4}{4} = 1;\]
\[(x - 0,5)(x - 1) \leq 0;\]
\[0,5 \leq x \leq 1;\]
\[Ответ:\ \ 0,5 \leq x \leq 1.\]
\[3)\ \left( \frac{13}{11} \right)^{x^{2} - 3x} < \frac{121}{169};\]
\[\left( \frac{13}{11} \right)^{x^{2} - 3x} < \left( \frac{11}{13} \right)^{2};\]
\[\left( \frac{13}{11} \right)^{x^{2} - 3x} < \left( \frac{13}{11} \right)^{- 2};\]
\[x^{2} - 3x < - 2;\]
\[x^{2} - 3x + 2 < 0;\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[x_{1} = \frac{3 - 1}{2} = 1\ \ и\ \]
\[\ x_{2} = \frac{3 + 1}{2} = 2;\]
\[(x - 1)(x - 2) < 0;\]
\[1 < x < 2;\]
\[Ответ:\ \ 1 < x < 2.\]
\[4)\ \left( 2\frac{2}{3} \right)^{6x^{2} + x} \leq 7\frac{1}{9};\]
\[\left( \frac{2 \bullet 3 + 2}{3} \right)^{6x^{2} + x} \leq \frac{7 \bullet 9 + 1}{9};\]
\[\left( \frac{8}{3} \right)^{6x^{2} + x} \leq \frac{64}{9};\]
\[\left( \frac{8}{3} \right)^{6x^{2} + x} \leq \left( \frac{8}{3} \right)^{2};\]
\[6x^{2} + x \leq 2;\]
\[6x^{2} + x - 2 \leq 0;\]
\[D = 1^{2} + 4 \bullet 6 \bullet 2 = 1 + 48 = 49\]
\[x_{1} = \frac{- 1 - 7}{2 \bullet 6} = - \frac{8}{12} = - \frac{2}{3};\]
\[x_{2} = \frac{- 1 + 7}{2 \bullet 6} = \frac{6}{2 \bullet 6} = \frac{1}{2};\]
\[\left( x + \frac{2}{3} \right)\left( x - \frac{1}{2} \right) \leq 0;\]
\[- \frac{2}{3} \leq x \leq \frac{1}{2};\]
\[Ответ:\ \ - \frac{2}{3} \leq x \leq \frac{1}{2}.\]
