ab=1
(a²+b²)/(a-b)
=[(a-b)²+2ab]/(a-b)
=[(a-b)²+2]/(a-b)
=(a-b) +2/(a-b)
由均值不等式得：(a-b)=2/(a-b)时,即a=b+√2时,(a-b)+2/(a-b)有最小值2√2
a->+∞时,b->0,(a-b)+2/(a-b)->+∞
综上得(a²+b²)/(a-b)的取值范围为[2√2,+∞)=[(a-b)²+2]/(a-b)=(a-b) +2/(a-b)上面一步如何化为下面一步....就是展开啊，(a-b)²/(a-b)=a-b2/(a-b)=2/(a-b)