$3 \sqrt{2} = 2$ ???

There are multiple ways to achieve this. The most obvious is $\frac{\sqrt{2} + \sqrt{2} }{ \sqrt{2} }$ . Another is $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$ . So the answer is obviously Yes.

yes first add two square root 2's and then divide by third

I was thinking of (\sqrt{2}^{\sqrt{2}})^\sqrt{2}=\sqrt{2}^2=2 .

$\bigg\lfloor\dfrac{\sqrt2}{\sqrt{2}}+\sqrt{2}\bigg\rfloor$ where $\lfloor \cdot \rfloor$ denotes the greatest integer function.

(2^1/2+2^1/2-2^1/2)^2 =2