已知f(2x+1)=x^2-2x，求f(x)与f(2x-1)的解析式

f(2x+1)=x^2-2x
=(4x^2+4x+1)/4-3x-1/4
=(2x+1)^2/4-(2x+1)*3/2+5/4
所以，
f(x)=x^2/4-3x/2+5/4

f(2x-1)=(2x-1)^2/4-3(2x-1)/2+5/4
=(4x^2-4x+1)/4-(6x-3)/2+5/4
=x^2-4x+3

设t = 2x + 1
x =（1-t）/2
f(t)=(1-t)^2/4 - (1-t)=(1-t)^2/4 + t-1
即f(x)=(1-x)^2/4 + x-1

f(2x-1)即把上面的t变成2x-1代入就可以求出了
f(2x-1)=(1-2x+1)^2/4 + 2x-1-1
=x^2+2x-2

已知f(2x+1)=x^2-2x，求f(x)与f(2x-1)的解析式

令2x+1=t
x=(t-1)/2
2x-1=t-2

f(t)=(t-1)^2/4-(t-1)=t^2/4-t/2+1/4-t+1=t^2/4-3t/2+5/4

f(t-2)=(t-3)^2/4-(t-3)=t^2/4-3t/2+9/4-t+3=t^2-5t/2+15/4

f(x)=x^2/4-3x/2+5/4
f(2x-1)=x^2-5x/2+15/4

f(2x-1)=(2x-1)^2-2(2x-1)=4x^2-8x+3

f(2x+1)=x^2-2x,令2x+1=t,x=(t-1)/2,则f(t)=(t^2)/4-6t/4+5/4,所以f(x)=(x^2)/4-6x/4+5/4.将2x-1带入即可!