Question

Question #03501

Answer 1

`(6x^2 - 4x)/(x^2 - 1)`

Explanation:

So, as your notation is not very clear to me, let me assume that you might have had this expression in mind:

`(x+1)/(x-1) + 3 / ((x+1)/(x-1)) + 2`

If this is the case, read on. If not, sorry. :-)

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Let's take care of the double fraction in the middle first.

If you divide a fraction by a fraction, you need to multiply with the reciprocal (your fraction "upside down") instead.

So, here:

`3 / ((x+1)/(x-1)) = 3 -: ((x+1)/(x-1)) = 3 * (x-1)/(x+1) = (3(x-1))/(x+1)`

Now, let's go on with the whole term:

`(x+1)/(x-1) + (3(x-1))/(x+1) + 2`

The smallest common multiple of your denominators `(x-1)` and `(x+1)` is `(x-1)(x+1)`. Thus, expand all the terms to bring them to this denominator:

`color(white)(xx) (x+1)/(x-1) + (3(x-1))/(x+1) + 2`

`= (x+1)/(x-1) * (x+1)/(x+1) + (3(x-1))/(x+1) * (x-1)/(x-1) + 2 * ((x+1)(x-1))/((x+1)(x-1))`

`= ((x+1)^2 + 3 (x-1)^2 + 2 (x+1)(x-1))/((x+1)(x-1))`

Now it's time to use the formula

`(a+b)^2 = a^2 + 2ab + b^2`

`(a-b)^2 = a^2 - 2ab - b^2`

`(a+b)(a-b) = a^2 - b^2`

Using those formula, we can simplify our expression as follows:

`... = (x^2 + 2x - 1 + 3(x^2 - 2x + 1) + 2 (x^2 -1))/(x^2 - 1)`

`color(white)(xu) = (6x^2 - 4x)/(x^2 - 1)`

or, if you prefer the completely factorized version,

`color(white)(xu) = (2x(3x-2))/((x+1)(x-1))`