An algebra problem by Satwik Murarka

$\text{Simplifying the denominator first}\\ 3\times(\sqrt{2+\sqrt{3}})=\frac{3\times(\sqrt{3}+1)}{\sqrt{2}}\\ \text{Multiplying numerator by √2}\\ 2\times(\sqrt{2}+\sqrt{6})=4\times(\sqrt{3}+1)\\ \text{Therefore, we get}\\ \frac{4\times(\sqrt{3}+1)}{3\times(\sqrt{3}+1)}=\frac{4}{3}\\ 4+3=\boxed{7}$

$\text{Let the given expression be equal to x}$ .

$\text{Then,}$ $x^2=\dfrac23\times\dfrac23\times\dfrac{(\sqrt2+\sqrt6)^2}{(2+\sqrt3)}$

$\text{So,}$ $x^2=\dfrac49\times\dfrac{(8+4\sqrt2)=4(2+\sqrt3)}{(2+\sqrt3)}=\dfrac49\times4$

$\text{Therefore,}$ $x=\sqrt{\left(\dfrac{4\times4}9\right)}=\dfrac43$

$\text{Finally we have got the fraction,already in its simplest form,thus,the answer is}$ $\text{4+3=7}$

$\begin{aligned} \dfrac{2 (\sqrt2 + \sqrt6)}{3\sqrt{2 + \sqrt3}} & = \dfrac{2 \sqrt2 (1 + \sqrt3)}{3 \sqrt{\frac{\sqrt3 + 1}{\sqrt3 -1}}} \\ \\ & = \dfrac{2\sqrt2 \sqrt{\sqrt3 + 1}}{\frac{3}{\sqrt{\sqrt3 -1}}} \\ \\ & = \dfrac{2\sqrt2 \sqrt{3-1}}{3} \\ \\ & = \dfrac{2\sqrt2 \times \sqrt2}{3} \\ \\ & = \dfrac{4}{3} = \dfrac{a}{b} \end{aligned} \\ \\ \therefore a =4 ~ \text{and} ~ b=3 \implies a+b = \boxed{7}$