Maths help please algebra

simplify expressing answer with a rational denominator.
1/√(x+1) - √(x-1)


FACTORISE
a(a+c)-b(b+c)

SIMPLIFY FULLY
(1/x-1) -(1/x+1)+1

1 / [√(x+1)-√(x-1)]
[1√(x+1)+√(x-1)] / [√(x+1)-√(x-1)][√(x+1)+√(x-1)]
[√(x+1)+√(x-1)] / [(x+1)+√(x-1)√(x+1) - √(x-1)√(x+1)-√(x-1)√(x-1)]
[√(x+1)+√(x-1)] / [x+1-(x-1)]
[√(x+1)+√(x-1)] / (x+1-x+1)
[√(x+1)+√(x-1)] / 2

a(a+c)-b(b+c)
a^2+ac-b^2-bc
a^2-b^2+ac-bc
(a+b)(a-b)+c(a-b)
(a-b)[(a+b)+c]
(a-b)(a+b+c)

[1/(x-1)] - [1/(x+1)] + 1
{1(x+1)/[(x-1)(x+1)]} - {1(x-1)/[(x+1)(x-1)]} + [1(x+1)(x-1)/[(x+1)(x-1)]}
{(x+1)/[(x-1)(x+1)]} - {(x-1)/[(x+1)(x-1)]} + {(x^2-1)/[(x+1)(x-1)]}
[(x+1)-(x-1)+(x^2-1)] / [(x+1)(x-1)]
(x+1-x+1+x^2-1) / [(x+1)(x-1)]
(x^2+1) / [(x+1)(x-1)]

EDIT: Okay, one of the lines wasn't showing up all the way because it was too long without spaces but I fixed it now.

wtf is the rational denominator

1. Don't feel like doing right now
2. a^2+ac-b^2-bc
3. 1 according to the way u wrote it