Question

Решите систему уравнений.

Answer 1

$left { {{x-y=sqrt3} atop {xy(x^2+y^2)=-1}} right.; left { {{(x-y)^2=3} atop {xy(x^2+y^2)=-1}} right. ; left { {{x^2+y^2-2xy=3} atop {xy(x^2+y^2)=-1}} right. ; left { {{x^2+y^2=3+2xy} atop {xy(3+2xy)=-1}} right.\\left { {{x^2+y^2=3+2xy} atop {3xy+2(xy)^2+1=0}} right. \\t=xy; ,; 2t^2+3t+1=0; ,; D=1; ,; t_1=-1; ,; t_2=-frac{1}{2}\\a); ; left { {{x-y=sqrt3} atop {xy=-1}} right. ; left { {{x=y+sqrt3; ; ; ; ; } atop {y^2+sqrt3y+1=0}} right.$

$y^2+sqrt3y+1=0; ,; ; D=3-4=-1<0; ; to; ; xin varnothing \\b); ; left { {{x=y+sqrt3} atop {xy=-frac{1}{2}}} right. left { {{x=y+sqrt3} atop {y^2+sqrt3y+frac{1}{2}=0}} right. \\y^2+sqrt3y+frac{1}{2}=0; ,; ; D=3-2=1; ,; y_1=frac{-sqrt3-1}{2}; ,; ; y_2=frac{-sqrt3+1}{2}\\x_1=frac{-sqrt3-1}{2}+sqrt3=frac{sqrt3-1}{2}; ; ,; ; x_2=frac{-sqrt3+1}{2}+sqrt3=frac{sqrt3+1}{2}\\Otvet:; ; (frac{sqrt3-1}{2},frac{-sqrt3-1}{2}); ,; (frac{sqrt3+1}{2},frac{-sqrt3+}{2}); .$