根号a(x-a)+根号a(y-a)=根号x-a-根号a-y,并且(a-x)(a-y)(x-y)不等于0,那么xx-xy+

Question

根号a(x-a)+根号a(y-a)=根号x-a-根号a-y,并且(a-x)(a-y)(x-y)不等于0,那么xx-xy+

Answer 1

√a〔x－a〕＋√a〔y－a〕=√x－a－√a－y
a[x-a]>=0 a[y-a]>=0 x-a>=0 a-y>=0 a,x,y是两两不等的实数
a[y-a]>=0 a-y>=0得：a=0
√a〔x－a〕＋√a〔y－a〕=√x－a－√a－y变为：
0=√x－√－y
x=-y
3x2＋xy－y2／x2－xy＋y2=(3x^2-x^2-x^2)/(x^2+x^2+x^2)=x^2/(3x^2)=1/3