1/[a+1]-[a+3]/[a²-1]×[a²-2a+1]/[a²+4a+3]
=1/[a+1]-[a+3]/[(a-1)(a+1)]×[(a-1)²]/[(a+1)(a+3)]
=[（a+1）-（a-1）]/[a+1]²
=2/[a+1]²
a²+2a-1=0
a²+2a+1=2
[a+1]²=2
所以1/[a+1]-[a+3]/[a²-1]×[a²-2a+1]/[a²+4a+3]=1