Question

Решите систему уравнений.

Answer 1

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

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