解由若f(x)=lg(m/(x-1)-1)图象关于原点中心对称
知f(x)是奇函数
故f(-x)=-f(x)
故f(-x)+f(x)=0
即lg(m/(-x-1)-1)+lg(m/(x-1)-1)=0
即lg(m/(-x-1)-1)(m/(x-1)-1)=0
即(m/(-x-1)-1)(m/(x-1)-1)=10^0
即(m/(-x-1)-1)(m/(x-1)-1)=1
即-(m/(x+1)+1)(m/(x-1)-1)=1
即(m/(x+1)+1)(m/(x-1)-1)=-1
即m^2/(x+1)(x-1)-m/(x+1)+m/(x-1)-1=-1
即m^2/(x+1)(x-1)-m/(x+1)+m/(x-1)=0
即m/(x+1)(x-1)-1/(x+1)+1/(x-1)=0
即m/(x+1)(x-1)=1/(x+1)-1/(x-1)
即m/(x+1)(x-1)=[(x-1)-(x+1)]/(x+1)(x-1)
即m/(x+1)(x-1)=-2/(x+1)(x-1).该式恒成立
故m=-2