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
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.
= = = = = = = = =
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.
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