Without loss of generality, let's assume $x > 0$. Then $$\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5=\left(\frac{\sqrt{x}}{\sqrt{2}}+\frac{1}{\sqrt{x}}\right)^{10}$$

Every term in its expansion can be written as $$\binom{10}{k}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)^k\left(\frac{1}{\sqrt{x}}\right)^{10-k}$$

It will become a constant when $10-k=k\implies k=5$. Accordingly, the constant term is $$\binom{10}{5}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)^5\left(\frac{1}{\sqrt{x}}\right)^{10-5}=\boxed{\frac{63\sqrt{2}}{2}}$$