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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 925
Задание 925
\[y = \log_{3|x| + 1}\sqrt{4x + 3}:\]
\[4y - \frac{1}{y} > 0\]
\[\frac{4y^{2} - 1}{y} > 0\]
\[\frac{(2y + 1)(2y - 1)}{y} > 0\]
\[- \frac{1}{2} < y < 0;\ \ \ y > \frac{1}{2}.\]
\[Область\ определения:\]
\[x > - \frac{3}{4};\ \ \ x \neq - \frac{1}{2};\ \ \ x \neq 0.\]
\[1)\ \sqrt{4x + 3} < 1\]
\[4x + 3 < 1\]
\[x < - \frac{1}{2}.\]
\[- \frac{1}{2} < \log_{3|x| + 1}\sqrt{4x + 3} < 0\]
\[\frac{1}{\sqrt{3|x| + 1}} < \sqrt{4x + 3} < 1\]
\[\sqrt{(4x + 3)\left( 3|x| + 1 \right)} > 1\ \ \ \]
\[(4x + 3)\left( 3|x| + 1 \right) > 1\ \ \ \]
\[12x|x| + 4x + 9|x| + 3 > 1\ \ \]
\[- 12x^{2} + 4x - 9x + 2 > 0\]
\[12x^{2} + 5x - 2 < 0\]
\[D = 25 + 96 = 121\]
\[x_{1} = \frac{- 5 - 11}{2 \bullet 12} = - \frac{2}{3};\]
\[x_{2} = \frac{- 5 + 11}{2 \bullet 12} = \frac{1}{4};\]
\[\left( x + \frac{2}{3} \right)\left( x - \frac{1}{4} \right) < 0\]
\[- \frac{2}{3} < x < \frac{1}{4}.\]
\[2)\ \log_{3|x| + 1}\sqrt{4x + 3} > \frac{1}{2}\]
\[\sqrt{4x + 3} > \sqrt{3|x| + 1}\]
\[4x + 3 > 3|x| + 1\]
\[4x + 3 > - 3x + 1\]
\[7x > - 2\]
\[x > - \frac{2}{7};\]
\[4x + 3 > 3x + 1\ \ \]
\[4x - 3x > - 2\ \ \]
\[x > - 2.\]
\[Ответ:\ \ \]
\[x \in \left( - \frac{2}{3};\ - \frac{1}{2} \right) \cup \left( - \frac{2}{7};\ 0 \right) \cup (0;\ + \infty).\]
\[y = \log_{2|x| + 1}(7x + 4):\]
\[y - \frac{1}{y} > 0\]
\[\frac{y^{2} - 1}{y} > 0\]
\[\frac{(y + 1)(y - 1)}{y} > 0\]
\[- 1 < y < 0;\ \ \ y > 1.\]
\[Область\ определения:\]
\[x > - \frac{4}{7};\ \ x \neq - \frac{3}{7};\ x \neq 0.\]
\[1)\ - 1 < \log_{2|x| + 1}(7x + 4) < 0\]
\[\frac{1}{2|x| + 1} < 7x + 4 < 1\]
\[(7x + 4)\left( 2|x| + 1 \right) > 1;\ x < - \frac{3}{7}\]
\[(7x + 4)(1 - 2x) > 1\]
\[7x - 14x^{2} + 4 - 8x > 1\]
\[14x^{2} + x - 3 < 0\]
\[D = 1 + 168 = 169\]
\[x_{1} = \frac{- 1 - 13}{2 \bullet 14} = - \frac{1}{2};\]
\[x_{2} = \frac{- 1 + 13}{2 \bullet 14} = \frac{3}{7};\]
\[\left( x + \frac{1}{2} \right)\left( x - \frac{3}{7} \right) < 0\]
\[- \frac{1}{2} < x < \frac{3}{7}.\]
\[2)\ \log_{2|x| + 1}(7x + 4) > 1\]
\[7x + 4 > 2|x| + 1\]
\[7x + 4 > 2x + 1\]
\[5x > - 3\]
\[x > - \frac{3}{5}.\]
\[7x + 4 > 1 - 2x\]
\[9x > - 3\]
\[x > - \frac{1}{3}.\]
\[Ответ:\ \]
\[x \in \left( - \frac{1}{2};\ - \frac{3}{7} \right) \cup \left( - \frac{1}{3};\ 0 \right) \cup (0;\ + \infty).\]
