分类：教育/科学-学习帮助
问题：一道高中函数题f(x)=lg(1+x/1
f(x)=lg(1+x/1-x),若f(y+z/1+yz)=1,f(y-z/1-yz)=2,-1<y<1,-1<z<1,
求f(y)、f(z)的值
回答：f[(y+z)/(1+yz)]

=lg{[1+(y+z)/(1+yz)]/[1-(y+z)/(1+yz)]}

=lg[(1+yz+y+z)/(1+yz-y-z)]

=lg[(1+y)(1+z)/(1-y)(1-z)]

=lg[(1+y)/(1-y)]+lg[(1+z)/(1-z)]

=f(y)+f(z)=1

f[(y-z)/(1-yz)]

=f{[1+(y-z)/(1-yz)]/[1-(y-z)/(1-yz)]}

=lg[(1-yz+y-z)/(1-yz-y+z)]

=lg[(1+y)(1-z)/(1-y)/(1+z)]

=lg[(1+y)/(1-y)+lg[(1-z)/(1+z)]

=f(y)-f(z)=2

f(y)=3/2; f(z)-1/2.