Simplifying and Rationalizing Problems?

Question

I'm having a trouble solving these two problems on my calculus summer assignment:

simplify: [ (1/x) - (1/5) ] / [ (1/(x^2)) - (1/25) ]

rationalize the denominator: 1 / ( 1 + √3 - √5 )

Please provide an explanation on how to approach and solve the problems... Thank you.

Answer 1

1/x^2 - 1/25 is a difference of two squares,
which equals (1/x - 1/5)(1/x + 1/5).

Now the numerator cancels with the left-hand term
and you are left with 1 / (1/x + 1/5), which equals
1 / [(x + 5) / (5x)]

= 5x / (x + 5)


1 / (1 + √3 - √5)
Think of the denominator as being 1 + (√3 - √5) and
multiply it by its 'conjugate' which is 1 - (√3 - √5).
And of course, multiply the numerator by this as well.
This gives you (1 - √3 + √5) / [1 - (√3 - √5)^2]
= (1 - √3 + √5) / (1 - 3 + 2√15 - 5)
= (1 - √3 + √5) / (2√15 - 7)
Now you can multiply by (2√15 + 7) / (2√15 + 7)
and your answer should be :
(2√15 + √5 + 3√3 + 7) / 11

Answer 2

simplify: [ (1/x) - (1/5) ] / [ (1/(x^2)) - (1/25) ]
1/x - 1/5 = (5-x)/5x
1/x^2 - 1/25 = (25-x^2)/25x^2
25-x^2 = (5-x)(5+x)
Invert and multiply (5-x)/5x * 25x^2/(5-x)(x+5) =
5x/(x+5)


rationalize the denominator: 1 / ( 1 + √3 - √5 )
Multiply top and bottom by (1 - √3 + √5)
(1- √3 + √5) / (1 -√3 + √5 +√3 - √9 + √15 -√5 +√15 - √25)=
(1 - √3 +√5) / (1 -3 - 5 + 2√15) =
(1 - √3 + √5) / (2√15 -7) =
Multiply top and bottom by 2√15 + 7
(2√15 -2√45 +2√75 +7 -7√3 +7√5) / (60 - 49) =
(2√15 - 6√5 +10√3 +7 -7√3 + 7√5) / 11 =
(2√15 +√5 +3√3 + 7) / 11