Rationalizing fractions with multiple radicals

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I am having trouble rationalizing $\frac{2}{\sqrt{2-\sqrt{2}}}$

I have tried multiplying the fraction by $\sqrt{2 + \sqrt{2}}$ and got $\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}$ I am not sure if that is correct or not but I then multiplied by $\sqrt{2}$ but got stuck...

Assuming that I did the problem correctly so far, how would I multiply $2\sqrt{2+\sqrt{2}}$ with $\sqrt{2}$ ?

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Instead of multiplying top and bottom by $\sqrt{2}$, note that $2=\left(\sqrt{2}\right)^2$, so that $$\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}=\sqrt{2}\sqrt{2+\sqrt{2}}=\sqrt{4+2\sqrt{2}}.$$