The maximum value of the term independent of $$t$$ in the expansion of $$\left(tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t}\right)^{10}$$ where $$x \in (0, 1)$$ is:

The general term in the expansion of $$\left(tx^{1/5} + \frac{(1-x)^{1/10}}{t}\right)^{10}$$ is $$\binom{10}{r} (tx^{1/5})^{10-r} \left(\frac{(1-x)^{1/10}}{t}\right)^r = \binom{10}{r} t^{10-2r} x^{(10-r)/5} (1-x)^{r/10}$$.

For the term independent of $$t$$, we need $$10 - 2r = 0$$, so $$r = 5$$. The term becomes $$\binom{10}{5} x (1-x)^{1/2} = 252 \cdot x(1-x)^{1/2}$$.

To maximize $$f(x) = x(1-x)^{1/2}$$ on $$(0,1)$$, we differentiate: $$f'(x) = (1-x)^{1/2} - \frac{x}{2(1-x)^{1/2}} = \frac{2(1-x) - x}{2(1-x)^{1/2}} = \frac{2 - 3x}{2(1-x)^{1/2}}$$.

Setting $$f'(x) = 0$$ gives $$x = \frac{2}{3}$$. The maximum value is $$f\left(\frac{2}{3}\right) = \frac{2}{3}\left(\frac{1}{3}\right)^{1/2} = \frac{2}{3\sqrt{3}}$$.

Therefore the maximum value of the term independent of $$t$$ is $$252 \cdot \frac{2}{3\sqrt{3}} = \frac{504}{3\sqrt{3}} = \frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$$.