Question

当x＞1时,求y＝（2x²－2x＋1）/（x－1）的最小值

Answer 1

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