计算题哈,若代数式x^2+y^2-14x+2y+50的值为0.化简求值：[2x^2-(x+y)(x-y)]-[(-x-y

如果使代数式x^2+y^2-14x+2y+50的值为0,则
x^2+y^2-14x+2y+50=0
(x-7)^2+(y+1)^2=0
x=7,y=-1
[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
=[2x^2-(x^2-y^2)]-[(x^2-y^2)+2y^2]
=(x^2+y^2)-(x^2+y^2)
=x^2+y^2-x^2-y^2
=0
(好像不应该是这样的呀,原题目中哪儿搞错了?)