1-(X+h) - 1-X
2+(X-H) 2-X
First of all, are you using h and H to represent the same thing? If so, be consistent, either use all "h" or all "H".
Next, is this your expression? $\displaystyle \displaystyle \begin{align*} \frac{1 - (x + h)}{2 + (x - h)} - \frac{1 - x}{2 - x} \end{align*}$?
Learning a little LaTeX would help everyone...
you keep changing signs in subsequent posts ... I get the feeling that this original expression has more than one sign typo.
is this what you originally meant ???
$\displaystyle \frac{1-(x+h)}{2+(x+h)} - \frac{1-x}{2+x}$
if not, fix it so that maybe someone can provide you with meaningful guidance.
The work you showed in your last post is a good first step. Just keep going...
$\displaystyle \frac{1-(x+h)}{2+(x+h)} - \frac{1-x}{2+x}$
$\displaystyle =\frac{\left[1-(x+h)\right](2+x)}{\left[2+(x+h)\right](2+x)} - \frac{\left[2+(x+h)\right](1-x)}{\left[2+(x+h)\right](2+x)}$
$\displaystyle =\frac{\left[1-(x+h)\right](2+x)-\left[2+(x+h)\right](1-x)}{\left[2+(x+h)\right](2+x)}$
Are you able to do the multiplication?
That's not what I'm getting. It would be helpful if you showed the actual work you did to get that answer...
$\displaystyle \frac{\left[1-(x+h)\right](2+x)-\left[2+(x+h)\right](1-x)}{\left[2+(x+h)\right](2+x)}$
$\displaystyle =\frac{(2+x)-\left(x^2 + 2h + 2x + hx\right)-(2-2x)-\left(x - hx - x^2 + h\right)}{\left[2+(x+h)\right](2+x)}$
$\displaystyle =\frac{2+x-x^2-2h-2x-hx-2+2x-x+hx+x^2-h}{\left[2+(x+h)\right](2+x)}$
Now combine like terms.