Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 376

\[\boxed{\mathbf{376}.}\]

\[\left\{ \begin{matrix} 2x^{2} - xy - y^{2} - 10x - 8y - 12 = 0 \\ 2x^{2} + 3xy + y^{2} + x - y - 6 = 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ 2x^{2} - xy - y^{2} - 10x -\]

\[- 8y - 12 = 0\]

\[2x^{2} - 2xy - 12x + xy + 2x -\]

\[- 2y - y^{2} - 6y - 12 = 0\]

\[2x(x - y - 6) + x(y + 2) -\]

\[- y(y + 2) - 6(y + 2) = 0\]

\[2x(x - y - 6) +\]

\[+ (y + 2)(x - y - 6) = 0\]

\[(x - y - 6)(2x + y + 2) = 0\ \ \]

\[2)\ 2x^{2} + 3xy + y^{2} +\]

\[+ x - y - 6 = 0\ \]

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

\[+ y^{2} - 3y + 2y - 6 = 0\]

\[2x(x + y + 2) + x(y - 3) +\]

\[+ y(y - 3) + 2 \cdot (y - 3) = 0\]

\[2x(x + y + 2) +\]

\[+ (y - 3)(x + y + 2) = 0\]

\[(x + y + 2)(2x + y - 3) = 0\ \ \ \]

\[\left\{ \begin{matrix} (x - y - 6)(2x + y + 2) = 0 \\ (x + y + 2)(2x + y - 3) = 0 \\ \end{matrix} \right.\ \]

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

\[x + 0 = 0\]

\[x = 0.\]

\[y = - x - 2 = - 2.\]

\[2)\ \left\{ \begin{matrix} x - y - 6 = 0\ \ \\ 2x + y - 3 = 0 \\ \end{matrix} \right.\ \ ( + )\]

\[3x = 9\]

\[x = 3.\]

\[y = x - 6 = 3 - 6 = - 3.\]

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

\[0 = - 5\]

\[нет\ корней.\]

\[4)\ \left\{ \begin{matrix} x - y - 6 = 0 \\ x + y + 2 = 0 \\ \end{matrix} \right.\ ( + )\]

\[2x = 4\]

\[x = 2.\]

\[y = - x - 2 = - 2 - 2 = - 4.\]

\[Ответ:(0;\ - 2);(3;\ - 3);(2; - 4).\]