已知x^2+4y^2-2x+4y+2=0求(xy-x^2)/(x^2-2xy+y^2)/xy的值
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已知x^2+4y^2-2x+4y+2=0求(xy-x^2)/(x^2-2xy+y^2)/xy的值
x^2+4y^2-2x+4y+2=0
(x^2-2x+1)+(4y^2+4y+1)=0
(x-1)^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-1=0,2y+1=0
x=1,y=-1/2
(xy-x^2)/[(x^2-2xy+y^2)/xy]
=xy(xy-x^2)/(x^2-2xy+y^2)
=xy*x(y-x)/(y-x)^2
=x^2y/(y-x)
=1^2*(-1/2)/(-1/2-1)
=(-1/2)/(-3/2)
=1/3
(x^2-2x+1)+(4y^2+4y+1)=0
(x-1)^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-1=0,2y+1=0
x=1,y=-1/2
(xy-x^2)/[(x^2-2xy+y^2)/xy]
=xy(xy-x^2)/(x^2-2xy+y^2)
=xy*x(y-x)/(y-x)^2
=x^2y/(y-x)
=1^2*(-1/2)/(-1/2-1)
=(-1/2)/(-3/2)
=1/3
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