已知f(x+1)=x^2－x,则f(x)=？

设x+1=t
所以x=t-1
带入得f(t)=(t-1)^2-(t-1)
=(t-1)(t-2)

所以f(x)=(x-1)(x-2)

f(x+1)=x^2－x
=(x+1)^2-3x-1
=(x+1)^2-3(x+1)+2

f(x)=x^2-3x+2

设u=x+1,x=u-1,代入方程得：f(u)=(u-1)^2-(u-1)=u^2-3u+2
则f(x)=x^2-3x+2

令a=x+1
则x=a-1
代入f(x+1)=x^2-x
f(a)=(a-1)^2-(a-1)=a^2-3a+2
所以f(x)=x^2-3a+2