Question

已知x的平方+4y的平方-4x+4y+5=0,求，如图

Answer 1

解：∵x²+4y²-4x+4y+5=0
（x-2）²+（2y+1）²=0
∴x=2，y=-1/2
则，所求原式=[2*4-（-1/2）*4]/(2²+2×（-1/2）-2×（-1/2）²)x(-2+2x(-1/2)/[2×（-1/2）-（-1/2²）]÷（-2²-（-1/2）²）/(-1/2)
=-51/136

Answer 2

解：
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

Answer 3

-6/85