\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-\left(\sqrt{2}+1\right)\right)}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}+1\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}-1\right)}{-2\sqrt{2}.\sqrt{2}}\)
\(=\dfrac{\sqrt{6}-2-\sqrt{2}}{-4}=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)
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