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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 126
Задание 126
\[\boxed{\mathbf{126}.}\]
\[1)\ \left\{ \begin{matrix} x^{2} - 2y - 2 = 0 \\ x^{2} + y^{2} - 5 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x^{2} - 2y = 2 \\ x^{2} + y^{2} = 5 \\ \end{matrix} \right.\ ( - )\]
\[- 2y - y^{2} = - 3\]
\[y^{2} + 2y - 3 = 0\]
\[D_{1} = 1 + 3 = 4\]
\[y_{1} = - 1 + 2 = 1;\ \ \]
\[y_{2} = - 1 - 2 = - 3.\]
\[x^{2} = 5 - y^{2}\]
\[x^{2} = 5 - 1 = 4\]
\[x = \pm 2.\]
\[x^{2} = - 3 - 1 = - 4\]
\[нет\ корней.\]
\[Ответ:\ (2;1);( - 2;1).\]
\[2)\ \left\{ \begin{matrix} x^{2} - 2xy + y^{2} = 4 \\ xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\left\{ \begin{matrix} (x - y)^{2} = 4 \\ xy = 3\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\lbrack \begin{matrix} \left\{ \begin{matrix} x - y = 2 \\ xy = 3\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ } \\ \left\{ \begin{matrix} x - y = - 2 \\ xy = 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - y = 2 \\ xy = 3\ \ \ \ \ \\ \end{matrix}\ \right.\ \text{\ \ \ }\left\{ \begin{matrix} x = 2 + y\ \ \ \ \ \ \ \ \ \ \\ (2 + y) \cdot y = 3 \\ \end{matrix} \right.\ \]
\[y^{2} + 2y - 3 = 0\]
\[D_{1} = 1 + 3 = 4\]
\[y_{1} = - 1 + 2 = 1;\ \]
\[\ y_{2} = - 1 - 2 = - 3.\]
\[x_{1} = 2 + y = 2 + 1 = 3;\]
\[x_{2} = 2 + y = 2 - 3 = - 1.\]
\[2)\ \left\{ \begin{matrix} x - y = - 2 \\ xy = 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \\ (x + 2)x = 3 \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 3 = 0\]
\[D_{1} = 1 + 3 = 4\]
\[x_{1} = - 1 + 2 = 1;\ \ \]
\[\ x_{2} = - 1 - 2 = - 3.\]
\[y_{1} = x + 2 = 1 + 2 = 3;\]
\[y_{2} = x + 2 = - 3 + 2 = - 1.\]
\[Ответ:( - 1; - 3);(3;1);\]
\[(1;3);( - 3;\ - 1).\]
\[3)\ \left\{ \begin{matrix} x - y = 17\ \ \ \\ \sqrt{x} - \sqrt{y} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[x - y = 17\]
\[\underset{1}{\overset{\left( \sqrt{x} - \sqrt{y} \right)}{︸}}\left( \sqrt{x} + \sqrt{y} \right) = 17\]
\[\sqrt{x} + \sqrt{y} = 17\]
\[\left\{ \begin{matrix} \sqrt{x} + \sqrt{y} = 17 \\ \sqrt{x} - \sqrt{y} = 1\ \ \ \\ \end{matrix} \right.\ ( + )\]
\[2\sqrt{x} = 18\]
\[\sqrt{x} = 9\]
\[x = 81.\]
\[y = x - 17 = 81 - 17 = 64.\]
\[Ответ:(81;64).\]
\[4)\ \left\{ \begin{matrix} x - y = 40\ \ \ \ \ \\ \sqrt{x} + \sqrt{y} = 20 \\ \end{matrix} \right.\ \]
\[x - y = 40\]
\[\underset{20}{\overset{\left( \sqrt{x} + \sqrt{y} \right)}{︸}}\left( \sqrt{x} - \sqrt{y} \right) = 40\]
\[20 \cdot \left( \sqrt{x} - \sqrt{y} \right) = 40\]
\[\sqrt{x} - \sqrt{y} = 2.\]
\[\left\{ \begin{matrix} \sqrt{x} - \sqrt{y} = 2\ \ \\ \sqrt{x} + \sqrt{y} = 20 \\ \end{matrix} \right.\ ( + )\]
\[2\sqrt{x} = 22\]
\[\sqrt{x} = 11\]
\[x = 121.\]
\[y = x - 40 = 121 - 40 = 81.\]
\[Ответ:(121;81).\]
