作业帮f[(3x+2)/2x-1]=x^2-1,求:f(1),f(x)
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f[(3x+2)/2x-1]=x^2-1,求:f(1),f(x)
f[(3x+2)/2x-1]=x^2-1,求:f(1),f(x)
f[(3x+2)/2x-1]=x^2-1,求:f(1),f(x)
令(3x+2)/2x-1＝1
则x=-3
f(1)=(-3)^2-1
=8
令(3x+2)/2x-1＝y
则x=(-2-y)/(3-2y)
f(y)=[(-2-y)/(3-2y)]^2-1
f(x)=[(-2-x)/(3-2x)]^2-1
1、(3x+2)/2x-1=1时，x=-3
f(1)=(-3)^2-1
=8
2、令(3x+2)/2x-1=t
解得x=(t+2)/(2t-3)
f(t)=[(t+2)/(2t-3)]^2-1
(3x+2)/2x-1=1求得x=2，所以f（1）=3
(3x+2)/2x-1=1/X+1/2，设1/X+1/2=T，则x=2/(2T-1)
f(T)=x^2-1=4/(2T-1)^2-1
所以f（x）=4/(2x-1)^2-1
令(3x+2)/2x-1=u
得 x=(u+2)/(2u-3)
所以 f(u)=[(u+2)/(2u-3)]^2-1
即 f(x)=[(x+2)/(2x-3)]^2-1
将x=1带入f(x)=[(x+2)/(2x-3)]^2-1
得 f(1)=8
令3x+2/2x-1=t,解得：x=t+2/2t-3 ,f(t)=(t+2/2t-3)^2-1=-3t^2+16t-5/(2t-3)^2=-(t-5)(3t-1)/(2t-3)^2
所以：f(x)==-(x-5)(3x-1)/(2x-3)^2 ,f(1)=8
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