Question 152499
I think you mean like this.
{{{(a-1)/(a+1)+(a+1)/(a-1)}}}
To add fractions, you need a common denominator.
The lowest common denominator would be (a+1)(a-1). 
Multiply the first fraction by (a-1)/(a-1) and
the second fraction by (a+1)(a+1) to achieve that.
{{{(a-1)/(a+1)+(a+1)/(a-1)=((a-1)(a-1))/((a+1)(a-1))+((a+1)(a+1))/((a-1)(a+1))}}}
{{{(a-1)/(a+1)+(a+1)/(a-1)=((a-1)(a-1)+(a+1)(a+1))/((a-1)(a+1))}}}
{{{(a-1)/(a+1)+(a+1)/(a-1)=((a^2-2a+1)+(a^2+2a+1))/((a-1)(a+1))}}}
{{{(a-1)/(a+1)+(a+1)/(a-1)=(2a^2+2)/((a-1)(a+1))}}}
{{{(a-1)/(a+1)+(a+1)/(a-1)=(2(a^2+1))/((a-1)(a+1))}}}