1. b/b-1 + 2b/b^2 -1

2. a-1/a+1 + a+1/a-1

2. Originally Posted by cutiepi_86
1. b/b-1 + 2b/b^2 -1
I presume this is
b/(b - 1) + 2b/(b^2 - 1)

Like adding any other kind of fraction, you need a common denominator. So factor
b^2 - 1 = (b + 1)(b - 1)

The first fraction only has a b - 1 in its denominator, so we need to multiply top and bottom by b + 1:

b/(b - 1) * (b + 1)/(b + 1) + 2b/[(b + 1)(b - 1)]

= b(b + 1)/[(b + 1)(b - 1)] + 2b/[(b + 1)(b - 1)]

= [b(b + 1) + 2b]/[(b + 1)(b - 1)]

= [b^2 + b + 2b]/[(b + 1)(b - 1)]

= [b^2 + 3b]/[(b + 1)(b - 1)]

= [b(b + 3)]/[(b + 1)(b - 1)]

-Dan

3. Originally Posted by cutiepi_86

1. b/b-1 + 2b/b^2 -1

2. a-1/a+1 + a+1/a-1

Again, this is
(a - 1)/(a + 1) + (a + 1)/(a - 1)
right?

To get a common denominator, we need to multiply the first fraction by (a - 1)/(a - 1) and the second fraction by
(a + 1)/(a + 1):

(a - 1)/(a + 1) * (a - 1)/(a - 1) + (a + 1)/(a - 1) * (a + 1)/(a + 1)

= [(a - 1)^2]/[(a + 1)(a - 1)] + [(a + 1)^2]/[(a + 1)(a - 1)]

= [(a - 1)^2 + (a + 1)^2]/[(a + 1)(a - 1)]

= [a^2 - 2a + 1 + a^2 + 2a + 1]/[(a + 1)(a - 1)]

= [2a^2 + 2]/[(a + 1)(a - 1)]

= [2(a^2 + 1)]/[(a + 1)(a - 1)]

-Dan

4. Thanks so much Dan!