# Thread: Bleh need some help :(

1. ## Bleh need some help :(

$1\over(x+h)^2$ - $1\over x^2$

Not very good at that LaTeX sort of stuff but here is what im stuck with and have to simplify it.

Its also all over h but no idea how to get that there :/

Anyways any suggestions?

2. Originally Posted by Djaevel
$1\over(x+h)^2$ - $1\over x^2$

Not very good at that LaTeX sort of stuff but here is what im stuck with and have to simplify it.

Its also all over h but no idea how to get that there :/

Anyways any suggestions?
get a common denominator and combine the two fractions.

3. Originally Posted by Djaevel
$1\over(x+h)^2$ - $1\over x^2$

Not very good at that LaTeX sort of stuff but here is what im stuck with and have to simplify it.

Its also all over h but no idea how to get that there :/

Anyways any suggestions?
If it's all over h I would multiply that by $\frac{1}{h}$:

$\frac{1}{h}(\frac{1}{(x+h)^2} - \frac{1}{x^2})$

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

Ignoring that h for now:

$\frac{1}{x^2+2hx+h^2} - \frac{1}{x^2}$

Cross multiply:

$\frac{2x^2+2hx+h^2}{x^2(x^2+2hx+h^2)}$

I admit it's very messy though. What did you need to find?

4. See i dont understand how you got the top part? I mean i did get the bottom somehow :/

Everything on here is all from my university pre calculus math class, and i dont remember ANYTHING from math 12 so im having a pretty rough time.

Like i dont remember most of the information from the units i've already done

5. $\frac{1}{h} \left[\frac{1}{(x+h)^2} - \frac{1}{x^2}\right]$

$\frac{1}{h} \left[\frac{x^2}{x^2(x+h)^2} - \frac{(x+h)^2}{x^2(x+h)^2}\right]$

$\frac{1}{h} \left[\frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2}\right]$

$\frac{1}{h} \left[\frac{-2xh - h^2}{x^2(x+h)^2}\right]$

$\frac{1}{h} \left[\frac{-h(2x + h)}{x^2(x+h)^2}\right]$

$-\frac{2x + h}{x^2(x+h)^2}$

6. Thanks skeeter always a great help