f(x+3)=x²-x-1,求f(x)帮我个忙,实在不会了,用配凑法,

f(x+3)=x²-x-1,求f(x)帮我个忙,实在不会了,用配凑法,
f(x+3)=x²-x-1,求f(x)
帮我个忙,实在不会了,用配凑法,

f(x+3)=x²-x-1,求f(x)帮我个忙,实在不会了,用配凑法,
f(x+3)
=x²-x-1
=(x+3)²-7(x+3)+11
所以将x+3用x代替
得到f(x)=x²-7x+11

f(x+3)=x²-x-1,求f(x)
令k = x+3 x = k-3
代入 x²-x-1
= k²-6k+ 9 - k +3 -1
= k² -7k + 8
所以 f(x) = x² -7x + 8

f=k²-7x+8
解一元二次方程就OK了

令x=x-3.则
f(x)=(x-3)²-(x-3)-1=x²-7x+11.

因为f(x+3)=（x+3）²-7x-10
=（x+3）²-7（x+3）+11
所以f（x）=x²-7x+11

配凑法：f(x+3)=（x+3）²-7(x+3)+11
令t=x+3,则f(t)=t²-7t+11
因此，f(x)=x²-7x+11