已知x>y>0且xy=1.求x2+y2/x-y的最小值及此時xy的值
已知x>y>0且xy=1.求x2+y2/x-y的最小值及此時xy的值
數(shù)學(xué)人氣：777 ℃時間：2019-10-24 04:30:19
優(yōu)質(zhì)解答
x>y>0且xy＝1,則依基本不等式得：(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.故所求最小值為：2√2.此時,xy＝1且x-y＝2/(x-y),取正根,解得：x...
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