已知x的平方+4y的平方-4x+4y+5=0,求,如图

x²+4y²-4x+4y+5=0
(x²-4x+4)+(4y²+4y+1)=0
(x-2)²+(2y+1)²=0
x-2=0 且 2y+1=0
x=2,y=-1/2
[(x^4-y^4)/(x²+xy-2y²)]×[(x+2y)/(xy-y²)]÷[(-x²-y²)/y]²
={ (x²+y²)(x+y)(x-y)/[(x+2y)(x-y)] }×{ (x+2y)/[y(x-y)] }×[y²/(x²+y²)²]
=y(x+y)/[(x-y)(x²+y²)]
=(-1/2)×(2-1/2)/{ (2+1/2)×[2²+(-1/2)²)] }
=-(3/4)/(5/2×17/4)
=-(3/4)/(85/8)
=-3/4×8/85
=-6/85

x²+4y²-4x+4y+5=0 化为 (x-2)²+(2y+1)²=0 解得：x=2, y=-1/2