Question

数学化简【a+1分之1 + a-1分之1】除以a²-2a+1分之2a

Answer 1

{[1/(a+1)]+[1/(a-1)]}/[2a/(a²-2a+1)]
={[(a-1)+(a+1)]/[(a+1)(a-1)]/[2a/(a-1)²]
={2a/[(a+1)(a-1)]}×(a-1)²/(2a)
=(a-1)/(a+1)

Answer 2

=【（a+1）（a-1）分之2a】乘以2a分之（a-1）的平方
=(a+1)分之（a-1）

Answer 3

（a-1）/（a+1）

Answer 4

（a-1）/（a+1）

Answer 5

解
[1/(a+1)+1/(a-1)]÷2a/(a²-2a+1)
=[(a-1)/(a-1)(a+1)+(a+1)/(a+1)(a-1)]×[(a-1)²/2a]
=[(a-1+a+1)/(a+1)(a-1)]×[(a-1)²/2a]
=[2a/(a-1)(a+1)]×[[(a-1)²/2a]
=(a-1)/(a+1)