八年级数学题，在线等！！！紧急！！！

x^2+4y^2-2x+4y+5=0
(x^2-4x+4)+(4y^2+4y+1)=0
(x-2)^2+(2y+1)^2=0
平方大于等于0，相加等于0
若有一个大于0，则另一个小于0，不成立。
所以两个都等于0
所以x-2=0,2y+1=0
x=2,y=-1/2

[(x^4-y^4)/(2x^2+xy-y^2)]*[(2x-y)/(xy-y^2)]÷[(x^2+y^2)^2/y^2]
=(x^4-y^4)(2x-y)y^2/[(2x^2+xy-y^2)(xy-y^2)(x^2+y^2)^2]
=(x^2+y^2)(x+y)(x-y)(2x-y)y^2/[(2x-y)(x+y)y(x-y)(x^2+y^2)^2]
=y/(x^2+y^2)
=(-1/2)/(4+1/4)
=-2/17

什么乱七八糟的,^是什么符号?还有求什么的值?写清楚好不好.

(x-1)^2+(2y+1)=0

x=1 y=-1/2

(x^4-y^4)(2x-y)(xy-y^2)y^2/(2x^2+xy-y^2)(xy-y^2)(x^2+y^2)^2
=y^3(x^2+y^2)(x+y)(2x-y)(x-y)^2/[(x+y)(2x-y)y(x-y)(x^2+y^2)^2
=y^2(x-y)/(x^2+y^2)
=y^2(x-y)/(x^2+y^2)
=(1/4)(3/2)/[1+(1/4)]
=3/10