Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 358

\[\boxed{\mathbf{358}.}\]

\[1)\ \left\{ \begin{matrix} 2x^{2} - 2xy + x = - 9 \\ 2y - 3x = 1\ \ \ \ \ \ \ \ \ | \cdot x \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x^{2} - 2xy + x = - 9\ \\ - 3x^{2} + 2xy - x = 0\ \\ \end{matrix} \right.\ ( + )\]

\[- x^{2} = - 9\]

\[x^{2} = 9\]

\[x = \pm 3.\]

\[2y = 1 + 3x\]

\[y = 0,5 + 1,5x.\]

\[x = 3:\]

\[y = 0,5 + 4,5 = 5;\]

\[x = - 3:\]

\[y = 0,5 - 4,5 = - 4.\]

\[Ответ:(3;5);( - 3; - 4).\]

\[2)\ \left\{ \begin{matrix} x^{2} + 6xy + 8y^{2} = 91 \\ x + 3y - 10 = 0\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 10 - 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (10 - 3y)^{2} + 6y(10 - 3y) + 8y^{2} = 91 \\ \end{matrix} \right.\ \]

\[100 - 60y + 9y^{2} + 60y -\]

\[- 18y^{2} + 8y^{2} - 91 = 0\]

\[- y^{2} = - 9\]

\[y^{2} = 9\]

\[y = \pm 3.\]

\[y = 3:\]

\[x = 10 - 3y = 10 - 9 = 1;\]

\[y = - 3:\]

\[x = 10 - 3y = 10 + 9 = 19.\]

\[Ответ:(1;3);\ \ (19; - 3).\]

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

\[\ \left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 - y - 1)(y - 1) = 2 \\ \end{matrix} \right.\ \]

\[(4 - y)(y - 1) = 2\]

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

\[- y^{2} + 5y - 6 = 0\]

\[y^{2} - 5y + 6 = 0\]

\[y_{1} + y_{2} = 5;\ \ \ \ \ \ y_{1} \cdot y_{2} = 6\]

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

\[x_{1} = 5 - 3 = 2;\ \ \ x_{2} = 5 - 2 = 3.\]

\[Ответ:(2;3);(3;2).\]

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

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

\[(y + 1)(y + 1) = 1\]

\[y^{2} + 2y + 1 = 1\]

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

\[y(y + 2) = 0\]

\[y_{1} = 0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_{2} = - 2;\]

\[x_{1} = 0 + 3 = 3;\ \ \ \]

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

\[Ответ:(3;0);(1; - 2).\]