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已知x>y>0且xy=1.求x2+y2/x-y的最小值及此时xy的值
人气:351 ℃ 时间:2019-10-24 04:30:19
解答
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...
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