f(-x)=-f(x)
loga[(1+mx)/(-x-1)]=-loga[(1-mx)/(x-1)]
(1+mx)/(-x-1)=(x-1)/(1-mx)
m²x²-x²=0
m=±1
當(dāng)x=1時(shí),(1-mx)/(x-1)=-1,不成立
∴m=-1
f(x)=loga[(x+1)/(x-1)]
(a-2+1)/(a-2-1)=a
a²-4a+1=0
a=2±√3
∵a>1
∴a=2+√3
r-1=0
∴r=1r-1=0那個(gè)是怎么得出來(lái)的？∵a>1∴f(r)→+∝,(r+1)/(r-1)→+∝因此,r-1=0為什么(r+1)/(r-1)→+無(wú)窮而r-1就得等于0(r+1)/(r-1)=1+2/(r-1)上式的1忽略不計(jì)，2/(r-1)→+∝，則r-1=0