Question

помогите!!! если лень решать, натолкните хотя бы на мысль ​

Answer 1

$\left \{ {{xy=2} \atop {x^{4}+x^{2}+y^{4}+y^{2}=22}} \right.$

Преобразуем второе уравнение системы:

$(x^{4} +2x^{2}y^{2}+y^{4} )-2x^{2} y^{2}+(x^{2}+2xy+y^{2})-2xy=22\\\\(x^{2}+y^{2})^{2}-8+(x+y)^{2}-4=22\\\\(x^{2}+y^{2})^{2}+(x+y)^{2}=34\\\\((x+y)^{2}-2xy) ^{2}+(x+y)^{2}-34=0\\\\((x+y)^{2}-4 )^{2}+(x+y)^{2}-34=0 \\\\(x+y)^{4}-8(x+y)^{2}+16+(x+y)^{2}-34=0\\\\(x+y)^{4}-7(x+y)^{2}-18=0$

$(x+y)^{2} =m,m\geq0\\\\m^{2}-7m-18=0\\\\Teorema \ Vieta:\\\\m_{1} =9\\\\m_{2}=-3<0-neyd\\\\(x+y)^{2} =9\\\\\left[\begin{array}{ccc}x_{0}+y_{0}=3 \\x_{0}+y_{0}=-3 \end{array}\right\\\\Otvet:\boxed{-3}$