Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 654

\[\boxed{\mathbf{654}.}\]

\[\left\{ \begin{matrix} \sqrt{25 - x^{2}} - \sqrt{25 - y^{2}} = \sqrt{8}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \sqrt{25 - x^{2}} + \sqrt{25 - y^{2}} = \sqrt{16 + (x + y)^{2}} \\ \end{matrix} \right.\ \]

\[ОДЗ:\]

\[\left\{ \begin{matrix} 25 - x^{2} \geq 0 \\ 25 - y^{2} \geq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} \leq 25 \\ y^{2} \leq 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 5 \leq x \leq 5 \\ - 5 \leq y \leq 5 \\ \end{matrix} \right.\ \]

\[Заменим:a = \sqrt{25 - x^{2}};\ \]

\[\ b = {\sqrt{25 - y^{2}}}^{'}\]

\[c = x + y;\ \ d = x - y;\]

\[cd = b^{2} - a^{2}\]

\[c^{2} + d^{2} = 100 - 2 \cdot \left( a^{2} + b^{2} \right)\]

\[\left\{ \begin{matrix} a - b = \sqrt{8}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ a + b = \sqrt{16 + c^{2}} \\ \end{matrix} \right.\ ( + )\]

\[2a = \sqrt{8} + \sqrt{16 + c^{2}}\]

\[a = \frac{1}{2}\left( \sqrt{8} + \sqrt{16 + c^{2}} \right)\]

\[\left\{ \begin{matrix} a + b = \sqrt{16 + c^{2}} \\ a - b = \sqrt{8}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ ( - )\]

\[2b = \sqrt{16 + c^{2}} - \sqrt{8}\]

\[b = \frac{1}{2}\left( \sqrt{16 + c^{2}} - \sqrt{8} \right)\]

\(c^{2} + d^{2} = 100 - 2 \cdot\)

\[cd = \frac{1}{4} \cdot \left( \sqrt{16 + c^{2}} - \sqrt{8} \right)^{2} -\]

\[- \frac{1}{4}\left( \sqrt{16 + c^{2}} + \sqrt{8} \right)^{2} =\]

\[= \frac{1}{4} \cdot \left( - 4\sqrt{8 \cdot \left( 16 + c^{2} \right)} \right) =\]

\[= - \sqrt{8} \cdot \sqrt{16 + c^{2}}.\]

\[\left( \text{cd} \right)^{2} = 8 \cdot \left( 16 + c^{2} \right) =\]

\[= 8c^{2} + 128.\]

\[\left\{ \begin{matrix} c^{2} + d^{2} = 76 - c^{2} \\ c^{2}d^{2} = 8c^{2} + 128\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} d^{2} = 76 - 2c^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ c^{2}\left( 76 - 2c^{2} \right) = 8c^{2} + 128 \\ \end{matrix} \right.\ \]

\[76c^{2} - 2c^{4} = 8c^{2} + 128\]

\[2c^{4} - 68c^{2} + 128 = 0\ \ \ \ \ |\ :2\]

\[c^{4} - 34c^{2} + 64 = 0\]

\[c_{1}^{2} = 32;\ \ c_{2}^{2} = 2.\]

\[1)\ \left\{ \begin{matrix} c^{2} = 32 \\ d^{2} = 12 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} (x + y)^{2} = 32 \\ (x - y)^{2} = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = \pm 4\sqrt{2} \\ x - y = \pm 2\sqrt{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \pm \left( 2\sqrt{2} - \sqrt{3} \right) \\ y = \pm \left( 2\sqrt{2} + \sqrt{3} \right) \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} c^{2} = 2\ \ \\ d^{2} = 72 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x + y)^{2} = 2\ \ \ \\ (x - y)^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = \pm \sqrt{2}\text{\ \ } \\ x - y = \pm 6\sqrt{2} \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} x = \pm \frac{5}{2}\sqrt{2} \\ y = \pm \frac{7}{2}\sqrt{2} \\ \end{matrix} \right.\ \]

\[Проверкой\ убираем\ \]

\[посторонние\ корни.\]

\[Ответ:\left( \frac{5}{2}\sqrt{2};\ - \frac{7}{2}\sqrt{2} \right);\]

\[\left( - \frac{5}{2}\sqrt{2};\ \frac{7}{2}\sqrt{2} \right);\]

\[\left( - 2\sqrt{2} + \sqrt{3};\ - 2\sqrt{2} - \sqrt{3} \right);\]

\[\left( 2\sqrt{2} - \sqrt{3};2\sqrt{2} + \sqrt{3} \right).\]