Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 115

\[\boxed{\mathbf{115}.}\]

\[1)\ \left\{ \begin{matrix} 2x - y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - y^{2} + 4 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - (2x)^{2} + 4 = 0 \\ \end{matrix} \right.\ \]

\[3x^{2} - 4x^{2} + 4 = 0\]

\[- x^{2} = - 4\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[\left\{ \begin{matrix} x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = 2 \cdot 2 = 4 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 2 \\ y = - 4 \\ \end{matrix} \right.\ \]

\[Ответ:(2;4);\ \ ( - 2; - 4).\]

\[2)\ \left\{ \begin{matrix} x - 2y = 8\ \ \ \ \ \ \ \\ x^{2} + 2y^{2} = 22 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 8 + 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (8 + 2y)^{2} + 2y^{2} = 22 \\ \end{matrix} \right.\ \]

\[64 + 32y + 4y^{2} + 2y^{2} - 22 = 0\]

\[6y^{2} + 32y + 42 = 0\ \ \ |\ :2\]

\[3y^{2} + 16y + 21 = 0\]

\[D_{1} = 64 - 63 = 1\]

\[y_{1} = \frac{- 8 + 1}{3} = - \frac{7}{3};\ \ \]

\[y_{2} = \frac{- 8 - 1}{3} = - 3.\]

\[x_{1} = 8 + 2 \cdot ( - 3) = 2;\]

\[x_{2} = 8 + 2 \cdot \left( - \frac{7}{3} \right) = 8 - \frac{14}{3} =\]

\[= 8 - 4\frac{2}{3} = 3\frac{1}{3}.\]

\[Ответ:(2; - 3);\left( 3\frac{1}{3}; - 2\frac{1}{3} \right).\]

\[3)\ \left\{ \begin{matrix} x^{2} + y^{2} = 13 \\ xy + 6 = 0\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} xy = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2xy + y^{2} - 2xy = 13 \\ \end{matrix} \right.\ \]

\[(x + y)^{2} - 2 \cdot ( - 6) = 13\]

\[(x + y)^{2} + 12 = 13\]

\[(x + y)^{2} = 1\]

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

\[\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \\ (1 - y)y = - 6 \\ \end{matrix} \right.\ \]

\[y - y^{2} + 6 = 0\]

\[y^{2} - y - 6 = 0\]

\[y_{1} + y_{2} = 1;\ \ y_{1} \cdot y_{2} = - 6\]

\[y_{1} = 3;\ \ \ y_{2} = - 2.\]

\[x_{1} = 1 - 3 = - 2;\]

\[x_{2} = 1 + 2 = 3.\]

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

\[\left\{ \begin{matrix} x = - 1 - y\ \ \ \ \ \ \ \ \ \\ ( - 1 - y)y = - 6 \\ \end{matrix} \right.\ \]

\[- y - y^{2} + 6 = 0\]

\[y^{2} + y - 6 = 0\]

\[y_{1} + y_{2} = - 1;\ \ y_{1} \cdot y_{2} = - 6\]

\[y_{1} = - 3;\ \ \ \ y_{2} = 2.\]

\[x_{1} = - 1 + 3 = 2;\ \ \]

\[x_{2} = - 1 - 2 = - 3.\]

\[Ответ:( - 2;3);(3;\ - 2);\]

\[(2;\ - 3);( - 3;2).\]