当x＞1时,求y＝（2x²－2x＋1）/（x－1）的最小值
问题描述：
当x＞1时,求y＝（2x²－2x＋1）/（x－1）的最小值
答
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