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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 617
Задание 617
\[\boxed{\mathbf{617}.}\]
\[1)\ \left\{ \begin{matrix} \sqrt{\frac{2x - 1}{y + 2}} + \sqrt{\frac{y + 2}{2x - 1}} = 2 \\ x + y = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 12 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{\frac{2x - 1}{12 - x + 2}} + \sqrt{\frac{12 - x + 2}{2x - 1}} = 2 \\ \end{matrix} \right.\ \]
\[\sqrt{\frac{2x - 1}{14 - x}} + \sqrt{\frac{14 - x}{2x - 1}} = 2\]
\[Пусть\ t = \sqrt{\frac{2x - 1}{14 - x}} > 0\]
\[t + \frac{1}{t} = 2\]
\[t^{2} - 2t + 1 = 0\]
\[(t - 1)^{2} = 0\]
\[t = 1.\]
\[\sqrt{\frac{2x - 1}{14 - x}} = 1\]
\[ОДЗ:\ \ x \neq 14.\]
\[\frac{2x - 1}{14 - x} = 1\]
\[2x - 1 = 14 - x\]
\[3x = 15\]
\[x = 5.\]
\[y = 12 - x = 12 - 5 = 7.\]
\[Ответ:(5;7).\]
\[2)\ \left\{ \begin{matrix} \sqrt{x^{2} + 5} + \sqrt{y^{2} - 5} = 5 \\ x^{2} + y^{2} = 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y^{2} = 13 - x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \sqrt{x^{2} + 5} + \sqrt{13 - x^{2} - 5} = 5 \\ \end{matrix} \right.\ \]
\[\left( \sqrt{x^{2} + 5} + \sqrt{8 - x^{2}} \right)^{2} = 5^{2}\]
\[ОДЗ:\ \ \ x^{2} \neq - 5;\ \ x^{2} \neq 8.\]
\[x^{2} + 5 + 2\sqrt{\left( x^{2} + 5 \right)\left( 8 - x^{2} \right)} +\]
\[+ 8 - x^{2} = 25\]
\[2\sqrt{\left( x^{2} + 5 \right)\left( 8 - x^{2} \right)} = 12\]
\[\sqrt{\left( x^{2} + 5 \right)\left( 8 - x^{2} \right)} = 6\]
\[\left( x^{2} + 5 \right)\left( 8 - x^{2} \right) = 36\]
\[8x^{2} + 40 - x^{4} - 5x^{2} = 36\]
\[- x^{4} + 3x^{2} + 4 = 0\]
\[x^{4} - 3x^{2} - 4 = 0\]
\[x^{2} = t \geq 0:\]
\[t^{2} - 3t - 4 = 0\]
\[t_{1} + t_{2} = 3;\ \ \ \ t_{1} \cdot t_{2} = - 4\]
\[t_{1} = 4;\ \ \ \ \]
\[\ t_{2} = - 1\ (не\ подходит).\]
\[x^{2} = 4\]
\[x = \pm 2.\]
\[y^{2} = 13 - x^{2} = 13 - 4 = 9\]
\[y = \pm 3.\]
\[Ответ:( - 2;\ - 3);( - 2;3);\]
\[(2;\ - 3);(2;3).\]
