



当x>1时,求y=2x^2-2x+1/x-1的最小值?

学习时间:2026-08-11 08:08:21 阅读：9589

当x>1时,求y=2x^2-2x+1/x-1的最小值?

最佳回答

眯眯眼的柚子

合适的羽毛

2026-08-11 08:08:21



y=[2x(x-1)+1]/(x-1)=2x(x-1)/(x-1)+1/(x-1)=2x+1/(x-1)=2(x-1)+1/(x-1)+2x>1,x-1>0所以2(x-1)+1/(x-1)>=2√[2(x-1)*1/(x-1)]=2√2所以y>=2√2+2所以最小值=2√2+2

最新回答共有2条回答

怕孤独的夕阳
回复
2026-08-11 08:08:21
y=[2x(x-1)+1]/(x-1)=2x(x-1)/(x-1)+1/(x-1)=2x+1/(x-1)=2(x-1)+1/(x-1)+2x>1,x-1>0所以2(x-1)+1/(x-1)>=2√[2(x-1)*1/(x-1)]=2√2所以y>=2√2+2所以最小值=2√2+2

热门文章

猜你喜欢