f(x)+ f(-x)=lg[x+√(x²+1)] +lg[√(x²+1)-x]=lg([x+√(x²+1)] [√(x²+1)-x] )=lg1=0即f(-x)=-f(x)f(-a)=-f(a)=-b注：f(x)=lg[x+√(x²+1)]因为√(x²+1)＞√(x²﹚=|x|∴x+√(x
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