f(2x-1)=x²+8所以换元,令2x-1=t,则x=(t+1)/2f(t)=x²+8=[(t+1)/2]²+8=[(t²+2t+1)/4]+8=(1/4)t²+(1/2)t+33/4再换回x,即可得到f(x)=(1/4)x²+(1/2)x+33/4