已知（x^2-2xy+y^2)+(-2x+2y)+1=0,求x-y的值

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已知（x^2-2xy+y^2)+(-2x+2y)+1=0,求x-y的值
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已知（x^2-2xy+y^2)+(-2x+2y)+1=0,求x-y的值
已知（x^2-2xy+y^2)+(-2x+2y)+1=0,求x-y的值

已知（x^2-2xy+y^2)+(-2x+2y)+1=0,求x-y的值
(x-y)²-2(x-y)+1=0
(x-y-1)²=0
x-y-1=0
x-y=1

（x^2-2xy+y^2)+(-2x+2y)+1=0
(x-y)²-2(x-y)+1=0
(x-y-1)²=0
x-y-1=0
x-y=1

解原等式可化为
（x-y）^2-2(x-y)+1=0
设x-y=a
则 a^2-2a+1=0
(a-1)^2=0
a=1
代入x-y=a可得 x-y=1

(x-y)^2-2(x-y)+1=(x-y-1)^2=0
则x-y-1=0
x-y=1

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