ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 653

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\[1)\ \left\{ \begin{matrix} x^{2} + y^{2} + xy = 84 \\ x + y + \sqrt{\text{xy}} = 14\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[Заменим:a = \sqrt{\text{xy}};\ \ b = x + y;\]

\[b^{2} = (x + y)^{2} = x^{2} + 2xy +\]

\[+ y^{2} = x^{2} + y^{2} + 2a^{2};\]

\[x^{2} + y^{2} = b^{2} - a^{2}.\]

\[\left\{ \begin{matrix} b^{2} - 2a^{2} + a^{2} = 84 \\ b + a = 14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b^{2} - a^{2} = 84 \\ b + a = 14\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b + a = 14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (b - a)(b + a) = 84 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b + a = 14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 14 \cdot (b - a) = 84\ \ \ |\ :14 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} b + a = 14 \\ b - a = 6\ \ \ \\ \end{matrix} \right.\ ( + )\]

\[2b = 20\]

\[b = 10.\]

\[a = 14 - b = 14 - 10 = 4.\]

\[\left\{ \begin{matrix} \left( \sqrt{\text{xy}} \right)^{2} = 4^{2} \\ x + y = 10\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = 10 - y \\ xy = 16\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 10 - y\ \ \ \ \ \ \ \ \\ y(10 - y) = 16 \\ \end{matrix} \right.\ \]

\[10y - y^{2} - 16 = 0\]

\[y^{2} - 10y + 16 = 0\]

\[D_{1} = 25 - 16 = 9\]

\[y_{1} = 5 + 3 = 8;\ \ \ \]

\[\text{\ \ }y_{2} = 5 - 3 = 2\]

\[x_{1} = 10 - 8 = 2;\ \ \]

\[\ x_{2} = 10 - 2 = 8.\]

\[Ответ:(2;8);(8;2).\]

\[2)\ \left\{ \begin{matrix} \sqrt{2x - 1} + \sqrt{3 - y} = 3 \\ 6x + y - 2xy = 7\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Заменим:a = \sqrt{2x - 1};\ \]

\[\ b = \sqrt{3 - y};\]

\[a^{2} \cdot b^{2} = (2x - 1)(3 - y) =\]

\[= 6x - 2xy - 3 + y\]

\[\left\{ \begin{matrix} \sqrt{2x - 1} + \sqrt{3 - y} = 3 \\ 6x + y - 2xy - 3 = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a + b = 3 \\ a^{2}b^{2} = 4\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} a = 3 - b\ \ \ \ \ \ \ \ \ \ \\ b^{2}(3 - b)^{2} = 4 \\ \end{matrix} \right.\ \]

\[b^{2}\left( 9 - 6b + b^{2} \right) = 4\]

\[b^{4} - 6b^{3} + 9b^{2} - 4 = 0\]

\[P(1) = P(2) = 0\]

\[(b - 1)(b - 2) = 0\]

\[b^{2} - 3b + 2 = 0\]

\[b^{2} - 3b - 2 = 0\]

\[D = 9 + 8 = 17\]

\[b = \frac{3 \pm \sqrt{17}}{2}.\]

\[1)\ b = 1:\]

\[a = 3 - b = 3 - 1 = 2.\]

\[\left\{ \begin{matrix} \left( \sqrt{3 - y} \right)^{2} = 1^{2}\text{\ \ } \\ \left( \sqrt{2x - 1} \right)^{2} = 2^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3 - y = 1\ \ \\ 2x - 1 = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} y = 2\ \ \ \\ x = 2,5 \\ \end{matrix} \right.\ \]

\[2)\ b = 2:\]

\[a = 3 - 2 = 1.\]

\[\left\{ \begin{matrix} \left( \sqrt{3 - y} \right)^{2} = 2^{2}\text{\ \ } \\ \left( \sqrt{2x - 1} \right)^{2} = 1^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3 - y = 4\ \ \\ 2x - 1 = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} y = - 1 \\ x = 1\ \ \ \\ \end{matrix} \right.\ \]

\[3)\ b = \frac{3 + \sqrt{17}}{2}:\]

\[a = 3 - \frac{3 + \sqrt{17}}{2}\]

\[нет\ решения.\]

\[4)\ b = \frac{3 - \sqrt{17}}{2}:\]

\[a = 3 - \frac{3 - \sqrt{17}}{2}\]

\[нет\ решения.\]

\[Ответ:(2,5;2);(1; - 1).\]