<s>[INST] 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.
 [/INST] $$\sqrt{3+2\sqrt2}=\sqrt{3+\sqrt8}=\sqrt\frac{3+\sqrt{9-8}}{2}+\sqrt\frac{3-\sqrt{9-8}}{2}=\sqrt2+1$$
$$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt12}=\sqrt\frac{4+\sqrt{16-12}}{2}-\sqrt\frac{4-\sqrt{16-12}}{2}=\sqrt3-1$$
Now we have:
$$\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=\sqrt2+2-\sqrt3$$
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