Question

решить систему уравнений (2х-5)(у+2)=0,  х^2+xy+y^2=4

Answer 1

$(2x-5)(y+2)=0\ x^2+xy+y^2=4\ \ 2xy+4x-5y-10=0\ x^2+xy+y^2=4\ \ x(2y+4)=5y+10\ x=frac{5y+10}{2y+4}\ \ (frac{5y+10}{2y+4})^2+(frac{5y+10}{2y+4})*y+y^2=4\ (5y+10)^2+(5y^2+10y)(2y+4)+y^2*(2y+4)^2=4*(2y+4)^2\ (y+2)^2(4y^2+10y+9)=0\ y=+-2\ x=0;2$