解：由f(1)=1得：1-m+n=-1m=n+2f(n)=mn²-mn+n=mm=n+2代入，得：n²-(n+2)n+n=n+22n=-2n=-1m=n+2=-1+2=1f(x)=x²-x-1f(-1)=(-1)²-(-1)-1=1f[f(-1)]=f(1)=1²-1-1=-1f[f(x)]=(x²-x-1)²-(x²-x-1)-1=x⁴-2x³-2x²+3x+1
