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
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
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
数学人气：392 ℃时间：2019-12-07 07:12:04
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