<s>[INST] Simplify an expression with a root $ \frac{\sqrt{-x+2}}{\sqrt{x-1}}$ Simplify an expression with a root 
I don't understand a transformation.
For this, $f_n= \frac{n}{2}l_n\sqrt{1-l_n^2/4}$
And now I'd like to show that
$\frac{1}{1-l_{2n}^2/4}f_{2n}=\frac{2}{1+\sqrt{1-l_n^2/4}}\cdot\frac{n}{2}l_n$
holds
Can anyone help me?
According to my calculations 
$f_{2n}=\frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^{2}/4}$ and then
$\frac{1}{1-l_{2n}^2/4}\frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^2/4}=n\frac{l_{2n}\sqrt{1-l_{2n}^2/4}}{1-l_{2n}^2/4}=n\cdot l_{2n}\cdot \frac{1}{\sqrt{1-l_{2n}^2/4}}=\frac{n}{2}\cdot l_{2n}\cdot \frac{2}{\sqrt{1-l_{2n}^2/4}}$
Also we have 
$l_{2n}^2=l_n^2/4+\big(1-\sqrt{1-l_n^2/4}\big)^2=2\big(1-\sqrt{1-l_n^2/4}\big)$
So I substituted the expression $l_{2n}$, but I can't prove the desired formula.
After putting in $n\cdot l_{2n}\cdot \frac{1}{\sqrt{1-l_{2n}^2/4}}$ I got 
$n\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)}\cdot \frac{1}{\sqrt{1-(2\big(1-\sqrt{1-l_n^2/4}\big))/4}}=n\frac{\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)}}=n\frac{\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)\sqrt{1-l_n^2/4}\big)}}{\sqrt{2\sqrt{1-l_n^2/4}-1}}=n\frac{\sqrt{-2\sqrt{1-l_n^2/4}+2}}{\sqrt{2\sqrt{1-l_n^2/4}-1}}$
Now I don't know how I can further simplify the Expression.
 [/INST] Nothing will be changed if you bring both the numerator and denominator under a single radical and bring the twos inside the radical to obtain
$$ n\sqrt{\frac{2-\sqrt{4-l_n^2}}{\sqrt{{4-l_n^2}}-1}} $$
Then the fraction under the radical can be rationalized.
$$ n\sqrt{\frac{2-\sqrt{4-l_n^2}}{\sqrt{{4-l_n^2}}-1}\cdot\frac{\sqrt{{4-l_n^2}}+1}{\sqrt{{4-l_n^2}}+1}}=n\sqrt{\frac{l_n^2-2+\sqrt{4-l_n^2}}{3-l_n^2}} $$
Although you could rationalize the denominator, I would not take it any further.
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