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Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$

The answer to this question is $\frac{4}{3}$ in a workbook.

How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2}$
$=$ $\sqrt{(\sqrt2)^2 + 2(\sqrt2)(1) + (1)^2}$
$=$ $\sqrt{(\sqrt2 + 1)^2}$
$=$ $\sqrt2 + 1$

But I can't simplify $\sqrt{2+\sqrt3}$  like that as $2+\sqrt3$ is can't be written as squares of two numbers. Is there any other method?

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Note that$$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)^2=\frac{4\left(8+4\sqrt3\right)}{9\left(2+\sqrt3\right)}=\frac{16}9=\left(\frac43\right)^2.$$

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Hints:

$$\sqrt 2+\sqrt 6=\sqrt 2(1+\sqrt 3)=\sqrt{2(1+\sqrt 3)^2}=\sqrt{2(4+2\sqrt 3)}=2\sqrt{2+\sqrt 3}$$

$$\sqrt{2+\sqrt 3}=\frac 1{\sqrt 2}\sqrt{4+2\sqrt 3}=\frac 1{\sqrt 2}\sqrt{(\sqrt 3+1)^2}=\frac{\sqrt 3+1}{\sqrt 2}=\frac{\sqrt 6+\sqrt 2}2$$

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$$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)= \frac{(2√2+2√6)\sqrt{2+\sqrt{3}}}{3(2+√3)}=\frac{8+4√3}{2+√3}=\frac{4}{3}.$$

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