已知x^2+4y^2-4x+4y+5=0,求(2x^2+xy-y^2分之x^4-y^4)乘以（xy-y^2分之2x-y)

x^2+4y^2-4x+4y+5=0,
(x-2)^2+(2y+1)^2=0,
x=2,y=-1/2,
(2x^2+xy-y^2分之x^4-y^4)*（xy-y^2分之2x-y)/[(x^2+y^2)/y]^2
=[(x-y)(x+y)(x^2+y^2)/(2x-y)(x+y)](2x-y)/([y(x-y)]*y^2/(x^2+y^2)^2
=y/(x^2+y^2)
=(-1/2)/(2^2+1/4)=-2/17.
x+3y+7z=0,3x-4y-18z=0,
x=2z,y=-3z,
（3x^2+5y^2+6z^2)/(x^2+2y^2+8z^2)
=(12z^2+45z^2+6z^2)/(4z^2+18z^2+8z^2)
=21/10.

^是什么计算符号呀？

x^2+4y^2-4x+4y+5=0
(x-2)^2+(2y+1)^2=0
(x-2)^2=0
x-2=0
x=2
(2y+1)^2=0
2y+1=0
y=-1/2