Question

x^2+4y^2-2x+4y+2=0求（xy-x^2)/(x^2-2xy+y^2)/xy的值

已知x^2+4y^2-2x+4y+2=0求（xy-x^2)/(x^2-2xy+y^2)/xy的值

Answer 1

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

Answer 2

x^2+4y^2-2x+4y+2=0
(x-1)^2+(2y+1)^2=0
x=1,y=-1/2
（xy-x^2)/(x^2-2xy+y^2)/xy
=x(y-x)/(x-y)^2/xy
=-1/y(x-y)=4/3

Answer 3

x² + 4y² - 2x + 4y + 2 = 0
x² - 2x + 1 + 4y² + 4y + 1 = 0
(x - 1)² + (2y + 1)² = 0
x = 1
y = -1/2
x + y = 1/2