已知x的平方+4y的平方-4x+4y+5=0,求,如图
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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衡改-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/咐弊判卜搏4)/(85/8)
=-3/4×8/85
=-6/85
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衡改-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/咐弊判卜搏4)/(85/8)
=-3/4×8/85
=-6/85
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