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解：1) (x+2a+1)/(x-3a+1)>0,=>a>0时，x>3a-1或x-2a-1或x f(x)=f(-x)，x=-x，则a无解；奇函数，=> f(x)+f(-x)=0,f(x)=lg[(x+2a+1)/(x-3a+1)]=lg[1+5a/(x-3a+1)],=> lg[1+5a/(x-3a+1)]+lg[1+5a/(-x-3a+1)]=0,=> [1+5a/(x-3a+1)]*[1+5a/(-x-3a+1)]=1,=> 1+5a/(x-3a+1)+5a/(-x-3a+1)+[5a/(x-3a+1)]*[1+5a/(-x-3a+1)]=1,=> ... |
