Question

Решите систему уравнений
x+y = 3,
{
х^2 +y^2 = 25 + 2ху.

Answer 1

$\begin{cases}x+y=3\\x^2+y^2=25+2xy\end{cases}begin{cases}x=3-y\\x^2+y^2=25+2xy\end{cases}begin{cases}x=3-y\\(3-y)^2+y^2=25+2y(3-y)\end{cases}$

$(3-y)^2+y^2=25+2y(3-y)9-6y+y^2+y^2=25+6y-2y^29-6y+2y^2=25+6y-2y^29-6y+2y^2-25-6y+2y^2=0-16-12y+4y^2=04y^2-12y-16=0\;|:4y^2-3y-4=0D=b^2-4ac=(-3)^2-4\cdot1\cdot(-4)=9+16=25y_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(-3)\pm\sqrt{25}}{2\cdot1}=\dfrac{3\pm5}{2}y_1=\dfrac{3-5}{2}=\dfrac{-2}{2}=-1y_2=\dfrac{3+5}{2}=\dfrac{8}{2}=4$

$x=3-yy_1=-1\;\Rightarrow\;x_1=3-y_1=3-(-1)=3+1=4y_2=4\;\Rightarrow\;x_2=3-y_2=3-4=-1$

ответ: (4; -1); (-1; 4).