已知实数a满足a^2+2a-8=0,求（a+1分之1）-（a^2-1分之a+3）乘以(a^2+4a+3分之a^2-2a+1)怎么做
问题描述：
已知实数a满足a^2+2a-8=0,求（a+1分之1）-（a^2-1分之a+3）乘以(a^2+4a+3分之a^2-2a+1)怎么做
答
原式=[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)]}
=[1/(a+1)]-[(a-1)/(a+1)²]
=[(a+1)/(a+1)²]-[(a-1)/(a+1)²]
=[(a+1)-(a-1)]/(a+1)²
=2/(a+1)²
答
a²+2a=8
原式=1/(a+1)-(a+3)/(a+1)(a-1)×(a-1)²/(a+1)(a+3)
=1/(a+1)-(a-1)/(a+1)²
=(a+1-a+1)/(a+1)²
=2/(a²+2a+1)
=2/(8+1)
=2/9