Find the value of $\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}$ for $x=\frac{\sqrt3}{4}$. Find the value of $$\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}$$ for $x=\dfrac{\sqrt3}{4}$.
I have no idea why I can't solve this problem. I tried to simplify the given expression, but I was able to reach only $$\dfrac{2-\sqrt{1+2x}\left(\sqrt{(1-2x)(1+2x)}-1\right)}{2x}$$ I also tried to plug in $x=\dfrac{\sqrt3}{4}$ directly, but I wasn't able to get anything. Thank  you!
 A: HINT...it may help you to use the fact that $$(1+\sqrt{3})^2=4+2\sqrt{3}$$
So $$\sqrt{1+\frac{\sqrt{3}}{2}}=?$$
Similarly with the $-$ sign
A: You can dodge a certain amount of work by re-writing the given expression using $ \ u \ = \ \sqrt{1 + 2x} \ $ and $ \ v \ = \ \sqrt{1 - 2x} \ \  $ to produce
$$  \frac{1 \ + \ 2x}{1 \ + \ \sqrt{1 + 2x}} \ + \ \frac{1 \ - \ 2x}{1-\sqrt{1-2x}} \ \ \rightarrow \ \ \frac{u^2}{1 \ + \ u} \ + \ \frac{v^2}{1 \ - \ v} \ \ = \ \ \frac{u^2 \ - \ u^2v \ + \ v^2 \ + \ uv^2 }{1 \ + \ u \ - \ v \ - \ uv}  $$ $$ = \ \ \frac{u^2 \ + \ v^2 \ - \ uv · (u \ - \ v)  }{1 \ + \ (u \ - \ v) \ - \ uv} \ \ . $$
We can now make a good deal of headway with $ \ u^2 \ = \ 1 + \frac{\sqrt3}{2}  \ \ , \ \ v^2 \ = \ 1 - \frac{\sqrt3}{2} \ \ ,  $ $ uv \ = \ \sqrt{1 - 4x^2} \ = \ \sqrt{1 - 4·\frac{3}{16}} \ = \ \sqrt{\frac14} \ = \ \frac12 \ \ . $
But we finally have to "buckle down" and deal with the difference $ \ u - v \ \ . $  We can make use of the fact that the irrational square-root under the radical remains part of the resulting value.  Thus, $ \ u \ = \ \sqrt{1 + \frac{\sqrt3}{2}} \ $ can be treated as $ \ \sqrt2·u \ = \ \sqrt{2 +  \sqrt3 } \ = \  a + b·\sqrt3  \ \ . $  This requires that $ \ a^2 + 3b^2 \ = \ 2 \ \ , \ \  2ab \ = \ 1 \ \ , \ $ for which we can obtain a solution "by inspection" (or more formally) as $ \ a \ = \ b \ = \ \frac{1}{\sqrt2} \ \ . $  From this, we obtain
$$ u \ \ = \ \  \sqrt{1 + \frac{\sqrt3}{2}} \ \ = \ \ \frac{\frac{1}{\sqrt2} + \frac{1}{\sqrt2}·\sqrt3 }{\sqrt2} \ \ = \ \ \frac12 + \frac{\sqrt3}{2}   \ \ \ , \ \ \ v \ \ = \ \  \sqrt{1 - \frac{\sqrt3}{2}} \ \  = \ \ -\frac12 + \frac{\sqrt3}{2}  $$
$$ \Rightarrow \ \ u \ - \ v \ \ = \ \ 1 \ \ . $$
(This sort of question comes up a lot on MSE: as just one example, here .)
Hence, for $ \ x \ = \ \frac{\sqrt3}{4} \ \ , $
$$  \frac{1 \ + \ 2x}{1 \ + \ \sqrt{1 + 2x}} \ + \ \frac{1 \ - \ 2x}{1-\sqrt{1-2x}} \ \ = \ \ \frac{\left(1 + \frac{\sqrt3}{2} \right) \ + \ \left(1 - \frac{\sqrt3}{2} \right) \ - \ \frac12 · 1 }{1 \ + \ 1 \ - \ \frac12} \ \ $$ $$ = \ \ \frac{1  \ + \ 1  \ - \ \frac12  }{\frac32} \ \ = \ \ 1 \ \ . $$
