对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值

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对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值对于函数f(x)=

对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值
对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(y
f(Y)和f(z)的值

对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值
注意到:
Ka=1+(y+z)/(1+yz)=(1+y+z+yz)/(1+yz)=(1+y)(1+z)/(1+yz)
Kb=1-(y+z)/(1+yz)=(1-y-z+yz)/(1+yz)=(1-y)(1-z)/(1+yz)
Kc=1+(y-z)/(1-yz)=(1+y-z-yz)/(1-yz)=(1+y)(1-z)/(1-yz)
Kd=1-(y-z)/(1-yz)=(1-y+z-yz)/(1-yz)=(1-y)(1+z)/(1-yz)
所以:
Ka/Kb=[(1+y)/(1-y)][(1+z)/(1-z)]
Kc/Kd=[(1+y)/(1-y)][(1-z)/(1+z)]
因为f(x)=lg[(1+x)/(1-x)]
所以f((y+z)/(1+yz))=lg(Ka/Kb)=lg[(1+y)/(1-y)][(1+z)/(1-z)]
=lg[(1+y)/(1-y)]+lg[(1+z)/(1-z)]
=f(y)+f(z)
同理f((y-z)/(1-yz))=f(y)-f(z)
解方程组:
f(y)+f(z)=1
f(y)-f(z)=2
可得:
f(y)=1.5
f(y)=-0.5