Question

若f（x+1）=x^2+2x+3，求f（x）的表达式。

Answer 1

令 x+1=t 则亮薯 x=t-1代入敬冲者判衡
f(t) = (t-1)^2 + 2(t-1) + 3
= t^2 -2t + 1 + 2t - 2 + 3
=t^2 + 2
所以f(x) = x^2 + 2

Answer 2

f(x)=x^2+2

设t=x+1 则此中f（森禅山t）=(t-1)^2+2(t-1)+3整理得袭粗f(t)=t^2+2,把t换成x，则f(x)=x^2+2

Answer 3

f(x+1)=(x+1-1)^2+2(x+1-1)+3替核兄蔽x+1为x得：
f(x)=(x-1)^2+2(x-1)+3整尘祥理得改州：
f(x)=x^2+2

Answer 4

f(x+1)=(x+1)^2+2
f(x)=x^2+2