Algebra review, please help?

Question

Can someone help me simplify these?

(1/x) - (1/5)

-----------------

(1/x^2) - (1/25)

I completely forgot how to do this.. DX

Also..

9 - x^(-2)

-------------

3 + x^(-1)

Answer 1

1) 1/x - 1/5    /      (1/x)^2  -  (1/5)^2

Now using the formula a^2 - b^2 = (a+b)(a-b)

1/x - 1/5  / (1/x - 1/5) (1/x + 1/5)

Therefore 1/x - 1/5 cancels out .. you are left with

(1)/(1/x + 1/5)

or

x + 5

because 1 is like saying 1/1 ...top multiplies the very bottom one and 2 middle ones multiply

2) Remember that you can make the exponents positive if you bring them down e.g.

9 - (1/x^2)       /     3 + 1/x

Now by common denominator, using separately for both numerator and denomination

[(9x^2 - 1) /x^2]   /  [(3x + 1)/x]

again the very top and the very bottom multiplies and the two mid ones

[9x^2 - 1 * x] / [(3x + 1)*x^2]

I believe you can take it from here

Goood luck :)      

Answer 2

GIVEN

(1/x) - (1/5)

-----------------

(1/x^2) - (1/25)

You can re-write the numerator as

(5 - x)/5x

and the denominator as

(25 - x^2)/25x^2

Hence,

(1/x) - (1/5)

-----------------

(1/x^2) - (1/25)



becomes equal to

(5 - x)/5x

-----------------

(25 - x)/25x^2

and can be simplified to

   x(5 - x)

-----------------

  (25 - x)^2

NOTE that the denominator in the above can be factored and so,

x (5 - x)

-----------------

(5 - x)(5 + x)

and the final simplified function is

      x

-----------------

  ( 5 + x)

Hope this helps.

Answer 3

[(1/x) - (1/5)]/[(1/x^2)  - (1/25)]

= [(5 - x)/5x]/[(25 - x^2)/25x^2]

= [(5 - x)/5x]/[(5 - x)(5 + x)/25x^2]

= 1/[(5 + x)/5x]

= 5x/(5 + x)

[9 - x^(-2)]/[3 + x^(-1)]

= [9 - 1/(x^2)]/[3 + 1/(x)]

= [(9x^2 - 1)/(x^2)]/[(3x + 1)/x]

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

= (3x - 1)/x