当x>1时,求y=(2x²-2x+1)/(x-1)的最小值
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当x>1时,求y=(2x²-2x+1)/(x-1)的最小值
当x>1时,求y=(2x²-2x+1)/(x-1)的最小值
当x>1时,求y=(2x²-2x+1)/(x-1)的最小值
y=[2(x²-2x+1)+2(x-1)+1]/(x-1)=[2(x-1)+1/(x-1)]+2
∵x>1∴x-1>0 2(x-1)>0 1/(x-1)>0
∴2(x-1)+1/(x-1)≥2√[2(x-1)×1/(x-1)]=2√2
(2(x-1)=1/(x-1)时,即x=1+√2/2时,
y取最小值2√2