Rational help, please!!!!?

Question

1.) x^2 + x / x^2 – 4 divided by x^2 – 1 / x^2 + 5x + 6
no.2) Find the least common denominator of the rational expressions: 7 / x^2 – 5x – 6 and x/ x^2 – 4x – 5
no.3) Simplify complex rational expression. 1 + 2/x
---------
1 – 4/x^2

Answer 1

[(x^2 + x) /( x^2 – 4)]/ [(x^2 – 1) /( x^2 + 5x + 6)]
= [(x(x+1))/(x+2)(x-2)][(x+2)(x+3)/(x+1)(x-1)]
x(x+3)/[(x-1)(x-2)], or
(x^2+3x)/(x^2-3x+2)

7 / x^2 – 5x – 6 and x/ x^2 – 4x – 5
= 7/[(x-6)(x+1)] and x/[(x-5)(x+1)]
So LCD = (x-6)(x-5)(x+1)

(1+2/x )/(1 - 4/x^2)
= [(x+2)/x]/ [(x^2-4)/x^2]
= [(x+2)/x][x^2/(x^2-4)]
= [(x+2)/x][x^2/(x-2)(x+2)]
= x/(x-2)

Answer 2

1) [ (x^2 + x) / (x^2 - 4) ] / [ (x^2 - 1) / (x^2 + 5x + 6) ]

The first step is to factor everything.

[ x(x + 1) / { (x - 2)(x + 2) } ] / [ { (x - 1)(x + 1) } / { (x + 5)(x + 1) } ]

We have fractions within a fraction; we don't want this. To fix this complex fraction, we must multiply top and bottom by
(x - 2)(x + 2)(x + 5)(x + 1). This gets rid of *all* fractions within the fraction, and it's important to know what is left behind.

[ x(x + 1)(x + 5)(x + 1) ] / [(x - 1)(x + 1)(x - 2)(x + 2)]

Now, we can cancel out one (x + 1), since it appears on top and bottom.

[ x(x + 5)(x + 1)] / [(x - 1)(x - 2)(x + 2)]

2) 7/(x^2 - 5x - 6) and x/(x^2 - 4x - 5)

To find the LCD, we factor the denominators.

7/[ (x - 6)(x + 1) ] and x/[(x - 5)(x + 1)]

As you can see, the LCD is (x - 6)(x + 1)(x - 5) {since they have a factor in common}.

3) (1 + 2/x) / (1 - 4/x^2)

This is a complex fraction; what we need to do is multiply top and bottom by the LCD of x and x^2; the LCD is x^2. This should get rid of *all* fractions.

Multiplying each term by x^2, we get

(x^2 + 2x) / (x^2 - 4)

Now, we factor everything.

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

And as you can see, the top and bottom have (x + 2) in common. Therefore, we can cancel them, leaving us with

x / (x - 2)

Answer 3

x=2