f(1-x/1+x)=1-x2/1+x2,则f(x)的解析式为帮帮忙 10
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用换元法思路很直接
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)
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)
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