∵a＜-b ∴a+b＜0,∴|a+b|=-(a+b) ∵-b＜1 ∴b+1>0,∴|b+1|=b+1 原式＝|a+b|/(b+a)-|a+b|/(b+1)/[(a+b)/|b+1|] =-(a+b)/(a+b)+(a+b)/(b+1)/[(a+b)/(b+1)] =-1+(a+b)(b+1)]/[(b+1)(a+b)] =-1+1 =0