Question

Решите систему уравнений:
х^2+y^2-6y=0
y+2x=0


И второе:
x^2-xy+y^2=63
x+y=-3

Answer 1

1)
$left { {{x^2+y^2-6y=0} atop {y+2x=0}} right. \y=-2x \x^2+(-2x)^2-6(-2x)=0 \x^2+4x^2+12x=0 \5x^2+12x=0 \x(5x+12)=0 \x_1=0 \5x+12=0 \5x=-12 \x_2= -frac{12}{5} =-2,4 \y_1=-2*0=0 \y_2=-2*-2,4=4,8$
Ответ: (0;0), (-2,4;4,8)
2)
$left { {{x^2-xy+y^2=63} atop {x+y=-3}} right. \ left { {{(x^2+2xy+y^2)-3xy=63} atop {x+y=-3}} right. \ left { {{(x+y)^2-3xy=63} atop {x+y=-3}} right. \left { {{(-3)^2-3xy=63} atop {x+y=-3}} right. \left { {{-3xy=63-9} atop {x+y=-3}} right. \left { {{xy=-18} atop {x+y=-3}} right. \x=-3-y \(-3-y)*y=-18 \y^2+3y=18 \y^2+3y-18=0 \D=9+72=81=9^2 \y_1= frac{-3+9}{2} =3 \y_2= frac{-12}{2} =-6 \x_1=-3-3=-6 \x_2=-3-(-6)=-3+6=3$
Ответ: (-6;3), (3;-6)