Найдите решение системы уравнений x+y+xy=5; x^2+y^2=5.

Ответ:

\[\left\{ \begin{matrix} x + y + xy = 5 \\ x^{2} + y^{2} = 5\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

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

\[x^{2} + 2xy + y^{2} = 5 + 10 - 2(x + y)\]

\[(x + y)^{2} = 15 - 2(x + y)\]

\[Пусть\ (x + y) = a:\]

\[a^{2} + 2a - 15 = 0\]

\[D = 1 + 15 = 16\]

\[a_{1} = - 1 + 4 = 3;\]

\[a_{2} = - 1 - 4 = - 5\ \]

\[1)\ a = - 5:\]

\[x + y = - 5\]

\[xy = 5 - (x + y) = 5 + 5 = 10.\]

\[t^{2} + 5t + 10 = 0\]

\[D = 25 - 40 < 0\]

\[нет\ корней.\]

\[2)\ a = 3:\]

\[x + y = 3\]

\[xy = 5 - (x + y) = 5 - 3 = 2\]

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

\[D = 9 - 8 = 1\]

\[t_{1} = \frac{3 + 1}{2} = 2;\ \ \]

\[t_{2} = \frac{3 - 1}{2} = 1\]

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

\[Ответ:(2;1);\ \ (1;2).\]