Question

化简：1/(1+1/a)+1/(1-1/a)/（1-1/a+1）/（1+1/a-1）

Answer 1

化简：[1/(1+1/a)+1/(1-1/a)]/[1-1/(a+1)]/[1+1/(a-1)]
解：原式=[a/(a+1)+a/(a-1)]÷(a/(a+1)÷(a/(a-1)
=[2a²/(a+1)(a-1)]×[(a+1)/a]×[(a-1)/a]
=[2a²/(a+1)(a-1)]×[(a+1)(a-1)/a²]
=2
①分子[1/(1+1/a)+1/(1-1/a)]上的中括号原题没有，是我加的，不知对不对？
②因为没有主分数线，因此无法判断是不是繁分数，只好当连除处理；若是繁分数，则应这样计算：
[a/(a+1)+a/(a-1)]/[(a/(a+1)÷(a/(a-1)]=[2a²/(a+1)(a-1)]/[(a-1)/a+1)]
=[2a²/(a+1)(a-1)]×[(a+1)/(a-1)]=2a²/(a-1)².

Answer 2

这是个连分式吗？层层算
（1-1/a+1）/（1+1/a-1）
=(a/(a+1))/(a/(a-1))
=(a-1)/(a+1)
1/(1-1/a)
=1/(a-1)
1/(1-1/a)/（1-1/a+1）/（1+1/a-1）
=1/(a-1)/[(a-1)/(a+1)]
=(a+1)/(a-1)^2
1/(1+1/a)=a/(a+1)
1/(1+1/a)+1/(1-1/a)/（1-1/a+1）/（1+1/a-1）
=a/(a+1)+(a+1)/(a-1)^2
=[a(a-1)^2-(a+1)^2]/[a(a-1)^2]
=(a^3-2a^2+a-a^2-2a-1)/[a(a-1)^2]
=(a^3-3a^2-a-1)/[a(a-1)^2]

Answer 3

（1-1/a+1）/（1+1/a-1）=(a-1)/(a+1);
原式等于:1/(1+1/a)+1/(1-1/a)/（a-1)/(a+1)=a[(a-1)^2+(a+1)^2]/（a+1)(a-1)^2=a(2a^2+2)/（a+1)(a-1)^2