解[1/(a+1)+1/(a-1)]÷2a/(a²-2a+1)=[(a-1)/(a-1)(a+1)+(a+1)/(a+1)(a-1)]×[(a-1)²/2a]=[(a-1+a+1)/(a+1)(a-1)]×[(a-1)²/2a]=[2a/(a-1)(a+1)]×[[(a-1)²/2a]=(a-1)/(a+1)