试说明代数式[(x-y)^2-(x+y)(x-y)]除(-2y)+y的值与y的值无关
问题描述：
试说明代数式[(x-y)^2-(x+y)(x-y)]除(-2y)+y的值与y的值无关
答
这题目不太对吧。。原题=
(-y)/[(x^2-2xy+y^2-(x^2-y^2)]=(-y)/(2y^2-2xy)=(-y)/2y(y-x)=-1/2（y-x）...
这...
答
[(x-y)^2-(x+y)(x-y)]除(-2y)+y
=(x²-2xy+y²-x²+y²)÷(-2y)+y
=(2y²-2xy)÷(-2y)+y
=-y+x+y
=x
∴结果与y的值无关
答
[(x-y)²-(x+y)(x-y)]除(-2y)+y
=[x²-2xy+y²-(x²-y²)]/(-2y)+y
=(-2xy+2y²)/(-2y)+y
=x-y+y
=x
所以代数式与y的值无关
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