先化简,再求值:1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/(a^2+4a+3),其中a满足a^2+2a-1=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/27 18:29:39
xQK
@Ό[;@H]@CrעMhYqkWVggYV9r^,a<6qG["#E1Q&-cXV5SևGumx#J|WbxQ C# BE2ʌ#@Ki !C
鍬̿nS=ef=EG3Z8"v<'vS
先化简,再求值:1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/(a^2+4a+3),其中a满足a^2+2a-1=0
先化简,再求值:1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/(a^2+4a+3),其中a满足a^2+2a-1=0
先化简,再求值:1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/(a^2+4a+3),其中a满足a^2+2a-1=0
1/(a+1)-[(a+3)/(a^2-1)]*[(a^2-2a+1)/(a^2+4a+3)]
=1/(a+1)-[(a+3)(a-1)^2]/[(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
又a^2+2a-1=0
a^2+2a+1=2
(a+1)^2=2
所以原式=2/(a+1)^2=2/2=1