Question

HEALP!!!!!!!!!!!!!!!Решите систему х+у+ху=-1=х^2+xy+y^2=3

Answer 1

$left { {{x+y+xy=-1} atop {x^2+xy+y^2=3}} right. ; left { {{x+y=-xy-1} atop {x^2+y^2=3-xy}} right. ; left { {{(x+y)^2=(-xy-1)^2} atop {x^2+2xy+y^2=3-xy+2xy}} right. \\ left { {{(x+y)^2=x^2y^2+2xy+1} atop {(x+y)^2=3+xy}} right. ; ; ; to ; ; ; x^2y^2+2xy+1=3+xy\\x^2y^2+xy-2=0\\t=xy; ,; ; ; t^2+t-2=0; ; to ; ; ; t_1=-2,; t_2=1; (Viet)$

$a); ; left { {{xy=-2} atop {x+y=-xy-1}} right. ; left { {{xy=-2} atop {x+y=1}} right. ; left { {{xy=-2} atop {y=1-x}} right.$

$x(1-x)=-2; ,; ; x-x^2=-2; ,; ; x^2-x-2=0\\x_1=-1; ,; ; x_2=2\\b); ; left { {{xy=1} atop {x+y=-2}} right. ; left { {{x(-x-2)=1} atop {y=-x-2}} right. ; ; to ; ; ; -x^2-2x=1\\x^2+2x+1=0; ,; ; (x+1)^2=0; ,; ; x=-1; ; to ; ; y=-1\\Otvet:; ; (-1,2); ,; ; (-1,-1); .$