y=(x²-2x+2)/(x-1)=[(x-1)²+1]/(x-1)=(x-1)+[1/(x-1)]
∵x∈(1,+∞)
∴x-1∈(0,+∞)
∴y=(x-1)+[1/(x-1)]≥2√{(x-1)*[1/(x-1)]}=2
∴y的最小值y(min)=2,当且仅当x-1=1/(x-1),且x>1,即x=2时取得