Question

СРОЧНО! решить систему уравнений

Answer 1

$\left \{ {{x^2-xy+y^2=3} \atop {(x+y)(x^2-xy+y^2)=9}} \right. \\ \\ \left \{ {{x^2-xy+y^2=3} \atop {(x+y)*3=9}} \right. \\ \\ \left \{ {{x^2-xy+y^2=3} \atop {x=3-y}} \right. \\ \\ (3-y)^2-y(3-y)+y^2-3 =0 \\ 9-6y+y^2-3y+y^2+y^2-3=0 \\ 3y^2-9y+6=0 \\ y^2-3y+2=0 \\ y=2; x =1 \\ y=1; x = 2$
Ответ: (1,2) и (2,1)