<s>[INST] Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$ 
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?



*

*I squared the whole denominator, but that didn't help.

*Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit.


Any help is appreciated.
 [/INST] The trick here is to multiply and divide by the conjugate.
For instance $$\frac{1}{1+\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{1-\sqrt{2}}{-1}$$
We can do the same here, but we will do it twice.
$$\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{((\sqrt{7}-2\sqrt{5})+\sqrt{3})((\sqrt{7}-2\sqrt{5})-\sqrt{3})} $$
$$=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{30-4\sqrt{35}}$$
Now multiply and divide by $30+4\sqrt{35}$ and you will elliminate the radical here as well.
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