Question

решить симметрическую систему уравнений:

$left { {{ x^{2} + xy + y^{2} = 3} atop {xy( x^{2} + y^{2}) = 2 }} right.$

Answer 1

$left { {{ x^{2} + xy + y^{2} = 3} atop {xy( x^{2} + y^{2}) = 2 }} right. \ x^2+y^2=a \ xy=b \ left { {{ a+b= 3} atop {ab= 2 }} right. \ a=3-b \ (3-b)b=2 \ b^2-3b+2=0 \ D=9-8=1 \ b_1=2 \ b_2=1 \ a_1=1 \ a_2=2$
$x^2+y^2=2 \ xy=1 \ x= frac{1}{y} \ frac{1}{y^2} +y^2-2=0 \ y^4-2y^2+1=0 \ (y^2-1)^2=0 \ y^2=1 \ y_1=1 \ y_2=-1 \ x_1=1 \ x_2=-1$
$x^2+y^2=1 \ xy=2 \ x= frac{2}{y} \ frac{4}{y^2} +y^2-2=0 \ 4y^4-2y^2+1=0 \ 4z^2-2z+1=0 \ D_1=1-4<0$
Ответ: (1; 1); (-1; -1)