Question

решите систему уравнений
$x {}^{2} + y {}^{2} = 4 \ xy = 1$

Answer 1

$left { {{x^2+y^2=4} atop {xy=1}} right. \\(x+y)^2=underbrace {x^2+y^2}_{4}+underbrace {2xy}_{2cdot 1}=4+2cdot 1=6; ; to ; ; left { {{x+y=pm sqrt6} atop {xy=1}} right. \\a); ; left { {{x+y=sqrt6} atop {xy=1}} right. ; left { {{y=sqrt6-x} atop {x(sqrt6-x)=1}} right. ; left { {{y=sqrt6-x} atop {x^2-xsqrt6+1=0}} right. \\x^2-sqrt6cdot x+1=0; ,; ; D=6-4=2; ,\\x_1= frac{sqrt6-sqrt2}{2}; ,; ; x_2= frac{sqrt6+sqrt2}{2}$

$y_1=sqrt6-frac{sqrt6-sqrt2}{2}=frac{sqrt6+sqrt2}{2}\\y_2=sqrt6-frac{sqrt6+sqrt2}{2}=frac{sqrt6-sqrt2}{2}\\b); ; left { {{x+y=-sqrt6} atop {xy=1}} right. ; left { {{y=-sqrt6-x} atop {x(-sqrt6-x)=1}} right. ; left { {{y=-sqrt6-x} atop {x^2+sqrt6cdot x+1=0}} right. \\x^2+sqrt6cdot x+1=0; ,; ; D=6-4=2\\x_1=frac{-sqrt6-sqrt2}{2}; ,; ; x_2=frac{-sqrt6+sqrt2}{2}\\y_1=-sqrt6-frac{-sqrt6-sqrt2}{2}=frac{-sqrt6+sqrt2}{2} \\y_2=-sqrt6-frac{-sqrt6+sqrt2}{2}=frac{-sqrt6+sqrt2}{2}$

$Otvet:; ; ( frac{sqrt6-sqrt2}{2};frac{sqrt6+sqrt2}{2}); ,; (frac{sqrt6+sqrt2}{2};frac{sqrt6-sqrt2}{2}); ,\\(frac{-sqrt6-sqrt2}{2}; frac{-sqrt6+sqrt2}{2}); ,; (frac{-sqrt6+sqrt2}{2}; frac{-sqrt6+sqrt2}{2}); .$