Question

如图，高一数学题，函数变形。

Answer 1

解：设t=(1-x)/(1+x)
则可得  x=(1-t)/(1+t)
把x代入
  f[(1-x)/(1+x)]=(1-x^2)/(1+x^2)中
可得
  f(t)={1-[(1-t)/(1+t)]^2}/{1+[(1-t)/(1+t)]^2}
      =[4t/(1+t)^2]/[(2t^2+2)/(1+t)^2]
      =2t/(1+t^2)
所以 f(x)=2x/(1+x^2)

Answer 2

令t=(1+x)/(1-x)
则x=(t-1)/(t+1)
代入原等式得：
f(t)=[1+(t-1)^2/(t+1)^2]/[1-(t-1)^2/(t+1)^2]
=[(t+1)^2+(t-1)^2]/[(t+1)^2-(t-1)^2]
=2(t^2+1)/(4t)
=(t^2+1)/(2t)
故f(x)=(x^2+1)/(2x)

Answer 3

解：换元法令t=(1+x)/(1-x)
则x=(t-1)/(t+1)
代入原等式得：
f(t)=[1+(t-1)^2/(t+1)^2]/[1-(t-1)^2/(t+1)^2]
=[(t+1)^2+(t-1)^2]/[(t+1)^2-(t-1)^2]
=2(t^2+1)/(4t)
=(t^2+1)/(2t)
即：f(x)=(x^2+1)/(2x)
希望对你有所帮助，还望采纳~~

Answer 4

换元发
令t=(1-x)/(1+x)
(1+x)t=1-x
tx+x=1-t
x=(1-t)/(1+t)
f(t)
=(1-x²)/(1+x²)
=[1-(1-t)²/(1+t)²]/[1+(1-t)²/(1+t)²]
=[(1+t)²-(1-t)²]/[(1+t)²+(1-t)²]
=[(1+2t+t²)-(1-2t+t²)]/[(1+2t+t²)+(1-2t+t²)]
=(4t)/(2+2t²)
=2t/(1+t²)
将t换回x，即得
f(x)=2x/(1+x²)

Answer 5

令t=（1+x）/（1-x)得x=(t-1)/（t+1)代入上式得f（t）=（t*t+1)/2t,再把t换成X就行