f(3x+1)=9x²-6x+5
=(9x²+6x+1)-(12x+4)+8
=(3x+1)²-4(3x+1)+8
令t=3x+1
则f(t)=f(3x+1)=t²-4t+8
即f(x)=x²-4x+8=(9x²+6x+1)-(12x+4)+8
=(3x+1)²-4(3x+1)+8
这两步是怎么得到的f(3x+1)=9x²-6x+5
=9x²-(12x-6x)+(1-4+8)
=9x²-12x+6x+1-4+8
=9x²+6x+1-12x-4+8
=(9x²+6x+1)-(12x+4)+8
=[(3x)²+2×3x+1²]-(4×3x+4×1)+8
=(3x+1)²-4(3x+1)+8