换元.可设t=x-1,由-1≤x≤3,===>-2≤x-1≤2,===>-2≤t≤2.f(x)=(x^2-5x+6)/(x^2+1)=(t^2-3t+2)/(t^2+2t+2)=1-[5t/(t^2+2t+2)].(1),当t=0时,f(t)=1,(2),当01-{5t/[t^2+2t+2]}≥1-[5(√2-1)/2]=(7-5√2)/2.===>(7-5√2)/2≤f(t)