先化简再求值2/a+1-a-2/a²-1除以a²-2a/a²-2a+1,其中a=5？
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2/a+1-a-2/a²-1除以a²-2a/a²-2a+1
=[2(a-1)-(a-2)](a-1)²除以(a+1)(a-1)a(a-2)
=a(a-1)²除以(a+1)(a-1)a(a-2)
=(a-1)除以(a+1)(a-2)
=4除以18
=2/9,7,[2/(a+1)-(a-2)/(a²-1)]÷(a²-2a)/(a²-2a+1)
=[2(a-1)-(a-2)]/[(a+1)(a-1)]÷[a(a-2)]/(a-1)²
=a/[(a+1)(a-1)]×(a-1)²/[a(a-2)]
=(a-1)/[(a+1)(a-2)]
=4/(6×3)
=4/18
=2/9
明白请采纳，有疑问请追问！
有新问题请求助，谢谢!,0,先化简再求值2/a+1-a-2/a²-1除以a²-2a/a²-2a+1,其中a=5
=[2(a-1)-(a-2)](a-1)²除以(a+1)(a-1)a(a-2)
=a(a-1)²除以(a+1)(a-1)a(a-2)
=(a-1)除以(a+1)(a-2)
=4除以18
=2/9,7,[2/(a+1)-(a-2)/(a²-1)]÷(a²-2a)/(a²-2a+1)
=[2(a-1)-(a-2)]/[(a+1)(a-1)]÷[a(a-2)]/(a-1)²
=a/[(a+1)(a-1)]×(a-1)²/[a(a-2)]
=(a-1)/[(a+1)(a-2)]
=4/(6×3)
=4/18
=2/9
明白请采纳，有疑问请追问！
有新问题请求助，谢谢!,0,先化简再求值2/a+1-a-2/a²-1除以a²-2a/a²-2a+1,其中a=5
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