已知f(1-x⼀1+x)=“1-x눀⼀1+x눀,则f(x)的解析式为()?

A:x/1+x눀B:-2x/1+x눀C:2x/1+x눀D:-x/1+x눀
2025-01-05 17:01:20
推荐回答(3个)
回答1:

设t=(1-x)/(1+x),则有t+tx=1-x, x=(1-t)/(1+t)
f(t)=[1-(1-t)^2/(1+t)^2]/[1+(1-t)^2/(1+t)^2]=[4t]/[2(1+t^2)=2t/(1+t^2)
故有f(x)=2x/(1+x^2)
选择C

回答2:

设u=1-x/1+x
u=1-2x/1+x
1-u=2x/1+x
1/1-u=1/2x+1/2
1/1-u-1/2=1/2x
u+1/2(1-u)=1/2x
x=1-u/1+u
f(u)=[(1+u)^2-(1-u)^2]/[(1+u)^2+(1-u)^2]
=4u/(2u^2+2)
=2u/(u^2+1)

C

回答3:

f(x)=2t/(!+t^2)