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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 616
Задание 616
\[\boxed{\mathbf{616}.}\]
\[1)\ \ \sqrt{\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4}} +\]
\[+ \sqrt{\frac{x^{2} + 2x + 4}{x^{2} - 2x + 3}} = \frac{5}{2}\]
\[\sqrt{\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4}} = t > 0:\]
\[t + \frac{1}{t} = \frac{5}{2}\ \ \ | \cdot 2t\]
\[2t^{2} + 2 = 5t\]
\[2t^{2} - 5t + 2 = 0\]
\[D = 25 - 16 = 9\]
\[t_{1} = \frac{5 + 3}{4} = 2;\ \ \]
\[\ t_{2} = \frac{5 - 3}{4} = \frac{1}{2}.\]
\[1)\ t_{1} = 2:\]
\[\sqrt{\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4}} = 2\]
\[\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4} = 4\]
\[\frac{x^{2} - 2x + 3 - 4 \cdot \left( x^{2} + 2x + 4 \right)}{x^{2} + 2x + 4} = 0\]
\[ОДЗ:\ x^{2} + 2x + 4 \neq 0\]
\[D_{1} = 1 - 4 = - 3 < 0\]
\[нет\ таких\ \text{x.}\]
\[x^{2} - 2x + 3 - 4x^{2} -\]
\[- 8x - 16 = 0\]
\[- 3x^{2} - 10x - 13 = 0\]
\[3x^{2} + 10x + 13 = 0\]
\[D = 100 - 156 = - 56 < 0\]
\[нет\ корней.\]
\[2)\ t_{2} = \frac{1}{2}:\]
\[\sqrt{\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4}} = \frac{1}{2}\]
\[\frac{x^{2} - 2x + 3}{x^{2} + 2x + 4} = \frac{1}{4}\]
\[4 \cdot \left( x^{2} - 2x + 3 \right) =\]
\[= x^{2} + 2x + 4\]
\[4x^{2} - 8x + 12 - x^{2} -\]
\[- 2x - 4 = 0\]
\[3x^{2} - 10x + 8 = 0\]
\[D_{1} = 25 - 24 = 1\]
\[x_{1} = \frac{5 + 1}{3} = 2;\ \ \ \]
\[x_{2} = \frac{5 - 1}{3} = \frac{4}{3}.\]
\[Ответ:x = \frac{4}{3};\ \ x = 2.\]
\[2)\ \sqrt{\frac{3x^{2} + x}{x^{2} - 1}} - \sqrt{\frac{x^{2} - 1}{3x^{2} + x}} = \frac{3}{2}\]
\[Пусть\ \sqrt{\frac{3x^{2} + x}{x^{2} - 1}} = t > 0:\]
\[t - \frac{1}{t} = \frac{3}{2}\ \ \ \ \ \ \ | \cdot 2t\]
\[2t^{2} - 2 - 3t = 0\]
\[2t^{2} - 3t - 2 = 0\]
\[D = 9 + 16 = 25\]
\[t_{1} = \frac{3 + 5}{4} = 2;\ \ \]
\[t_{2} = \frac{3 - 5}{4} =\]
\[= - \frac{1}{2}\ (не\ подходит).\]
\[t = 2:\]
\[\sqrt{\frac{3x^{2} + x}{x^{2} - 1}} = 2\]
\[\frac{3x^{2} + x}{x^{2} - 1} = 4\]
\[ОДЗ:\ \ \ x^{2} - 1 \neq 0;\ \ x \neq \pm 1.\]
\[3x^{2} + x = 4x^{2} - 4\]
\[x^{2} - x - 4 = 0\]
\[D = 1 + 16 = 17\]
\[x = \frac{1 \pm \sqrt{17}}{2}.\]
\[Ответ:x = \frac{1 \pm \sqrt{17}}{2}.\]
