y=(x^2+1)^2/4x
=(x^2+2x+1)/(4x)
=(x/4)+(1/2)+(1/4x)≥2√(x/4)*(1/4x)+(1/2)=(1/2)+(1/2)=1
當(dāng)x/4=1/4x時(shí),x=1
取得最小值：ymin=1
求y=(x^2+1)^2/4x的最小值,怎么求?
求y=(x^2+1)^2/4x的最小值,怎么求?
數(shù)學(xué)人氣：631 ℃時(shí)間：2020-06-16 08:33:33
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