1、设1+(1/x)=t,则：x=1/(1-t),1/x=1-t
∵f(1+(1/x))=x^2+(1/x^2)+(3/x)
=x^2+(1/x)[(1/x)+3]
∴f(t)=[1/(1-t)]^2+(1-t)[(1-t)+3]
=[1/(t-1)]^2+(t-1)(t-4)
∴f(x)=[(x-1)^3(x-4)+1]/(x-1)^2
2、设1+2√x=t,则：x=(t-1)^2/4,√x=(t-1)/2
又f(1+2√x)=2x+√x
∴f(t)=2[(t-1)^2/4]+[(t-1)/2]
=[(t-1)/2](t-1+1)
=t(t-1)/2
∴f(x)=x(x-1)/2