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当x>1时,求y=2x^2-2x+1/x-1的最小值?

学习时间:2026-05-14 02:13:08 阅读：9877

当x>1时,求y=2x^2-2x+1/x-1的最小值?

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2026-05-14 02:13:08

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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

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2026-05-14 02:13:08
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

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