作业帮已知：a²+2a-1=0,求1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)的值
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已知：a²+2a-1=0,求1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)的值
已知：a²+2a-1=0,求1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)的值
已知：a²+2a-1=0,求1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)的值
a²+2a-1=0
a²+2a=1
a²+2a+1=2
(a+1)²=2
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)]/(a+1)²
=2/(a+1)²
=2/2
=1
先对1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)进行化简
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...
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先对1/(a+1)-(a+3)/(a²-1)×（a²-2a+1)/(a²+4a+3)进行化简
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)]/(a+1)²
=2/(a+1)²
由于a²+2a-1=0，于是(a+1)²=a²+2a+1=2
于是原式=1
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