已知x满足2x2+4x+y2=2xy-4 求：xy

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已知x满足2x2+4x+y2=2xy-4 求：xy
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已知x满足2x2+4x+y2=2xy-4 求：xy
已知x满足2x2+4x+y2=2xy-4 求：xy

已知x满足2x2+4x+y2=2xy-4 求：xy
以下用 x^2 表示 x的平方.
= = = = = = = = =
因为 2x^2 +4x +y^2 =2xy -4,
所以 0 =2x^2 -2xy +y^2 +4x +4
=(x^2 -2xy +y^2) +(x^2 +4x +4)
=(x-y)^2 +(x+2)^2.
又因为 (x-y)^2 ≥0,
(x+2)^2 ≥0,
所以 (x-y)^2 =0,
(x+2)^2 =0.
即 x -y=0,
x +2 =0,
解得 x =y = -2.
所以 xy =4.
= = = = = = = = =
配方法.

2x2+4x+y2=2xy-4
(x^2-2xy+y^2)+(x^2+4x+4)=0
(x-y)^2+(x+2)^2=0
上式只有在x-y=0和x+2=0时，才能成立
所以x=y, x=-2
故xy=x^2=(-2)^2=4

XY=4

说清楚点，题好像不是很清楚啊！！！

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