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

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 11:35:59
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
x){}Klgض91H(Q@Ʀ{: v=m~O.H H*m |sƓK5@:M"P)*#^g }Ovv

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

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
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

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