simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But even with that I don't get that result.

We can recognize both expressions as squares: 


*

*$3+2\sqrt2=2+2\sqrt2+1=(\sqrt2)^2+2\cdot \sqrt2\cdot 1+1^2=(\sqrt2+1)^2$, and 

*$4-2\sqrt3=3-2\sqrt3+1=(\sqrt3-1)^2$.


This means that $\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=2+\sqrt2-\sqrt3$.
