已知5x^2-4xy+y^2-2x+1=0 求(x-y)^2012的值 .
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已知5x^2-4xy+y^2-2x+1=0 求(x-y)^2012的值 .
已知5x^2-4xy+y^2-2x+1=0 求(x-y)^2012的值 .
已知5x^2-4xy+y^2-2x+1=0 求(x-y)^2012的值 .
5x^2-4xy+y^2-2x+1=0
(4x^2-4xy+y^2)+(x^2-2x+1)
=0
(2x-y)^2+(x-1)^2=0
2x-y=0,x-1=0
x=1,y=2.
(x-y)^2012=(1-2)^2012=1
4X^2-4xy+y^2+x^2-2x+1=0
(2x-y)^2+(x-1)^2=0
所以2x-y=0且x-1=0
x=1
y=2
(x-y)^2012=1
5x^2-4xy+y^2-2x+1=0 化简后得: (2x-y)^2+(x-1)^2=0 即:(2x-y)^2= -(x-1)^2 则两边均为0
可求得x=1,y=2
(x-y)^2012=(1-2)^2012=1
由题目知,4x^2-4xy+y^2+x^2-2x+1=0,配方(2x-y)^2+(x-1)^2=0,
则得2x-y=0,x-1=0,所以x=1,y=2,故得,(x-y)^2012=1