已知x>y>0且xy=1.求x2+y2/x-y的最小值及此时xy的值

x>y>0且xy＝1,则依基本不等式得：(x²+y²)/(x-y)＝[(x-y)²+2xy]/(x-y)＝(x-y)+2xy/(x-y)＝(x-y)+2/(x-y)≥2√[(x-y)·2/(x-y)]＝2√2.故所求最小值为：2√2.此时,xy＝1且x-y＝2/(x-y),取正根,解得：x...