f(x)+f(-x)=loga[(1-mx)/(x-1)]+loga[-(1+mx)/(x+1)]
=loga[(mx+1)(mx-1)/(x+1)(x-1)]=loga[(m^2x^2-1)/(x^2-1)]=0
所以(m^2x^2-1)/(x^2-1)=1 m^2-1=0 m=1(舍去),m=-1 f(x)=loga[(1+2/(x-1)]
当a＞1,x属于（1,√3）时,f(x)单调递减 f(x)>f(√3)
f(√3)=loga[1+2/(√3-1)]=loga[√3+2]=1
a=√3+2