x>0，y>0，且xy＝1，求(x²+y²)/(x-y)的范围
1个回答
展开全部
(1)当x>y时
(x²＋y²﹚／﹙x-y)
=[(x-y)^2+2xy]/(x-y)
=(x-y)+2xy/(x-y)
≥2√[2(x-y)*2xy/(x-y)]=4
则有(x²＋y²﹚／﹙x-y﹚≥4
(2)当0<x<y时
x-y<0
(x²＋y²﹚／﹙x-y)
=[(x-y)^2+2xy]/(x-y)
=-[(y-x)+2xy/(y-x)]
(x²＋y²﹚／﹙x-y)≤-2√[2(y-x)*2xy/(y-x)]=4
x-y<0
则有(x²＋y²﹚／﹙x-y﹚<0
故 (x²+y²)/(x-y)<0或(x²+y²)/(x-y)≥4
(x²＋y²﹚／﹙x-y)
=[(x-y)^2+2xy]/(x-y)
=(x-y)+2xy/(x-y)
≥2√[2(x-y)*2xy/(x-y)]=4
则有(x²＋y²﹚／﹙x-y﹚≥4
(2)当0<x<y时
x-y<0
(x²＋y²﹚／﹙x-y)
=[(x-y)^2+2xy]/(x-y)
=-[(y-x)+2xy/(y-x)]
(x²＋y²﹚／﹙x-y)≤-2√[2(y-x)*2xy/(y-x)]=4
x-y<0
则有(x²＋y²﹚／﹙x-y﹚<0
故 (x²+y²)/(x-y)<0或(x²+y²)/(x-y)≥4
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
