The value $$9\int_0^9 \left[\sqrt{\frac{10x}{x+1}}\right] dx$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is _____.

We need $$9\int_0^9 \left[\sqrt{\frac{10x}{x+1}}\right] dx$$.

Let $$u = \sqrt{\frac{10x}{x+1}}$$. As $$x$$ goes from 0 to 9, $$u$$ goes from 0 to $$\sqrt{90/10} = 3$$.

We find where $$u = k$$ for integer $$k$$: $$\frac{10x}{x+1} = k^2 \Rightarrow 10x = k^2(x+1) \Rightarrow x(10-k^2) = k^2 \Rightarrow x = \frac{k^2}{10-k^2}$$.

For $$k = 0$$: $$x = 0$$. For $$k = 1$$: $$x = \frac{1}{9}$$. For $$k = 2$$: $$x = \frac{4}{6} = \frac{2}{3}$$. For $$k = 3$$: $$x = \frac{9}{1} = 9$$.

So $$[\sqrt{\frac{10x}{x+1}}] = 0$$ on $$[0, \frac{1}{9})$$, $$= 1$$ on $$[\frac{1}{9}, \frac{2}{3})$$, $$= 2$$ on $$[\frac{2}{3}, 9)$$, $$= 3$$ at $$x = 9$$.

$$\int_0^9 = 0 \cdot \frac{1}{9} + 1 \cdot \left(\frac{2}{3} - \frac{1}{9}\right) + 2 \cdot \left(9 - \frac{2}{3}\right) + 3 \cdot 0$$

$$= 0 + \frac{5}{9} + 2 \cdot \frac{25}{3} = \frac{5}{9} + \frac{50}{3} = \frac{5}{9} + \frac{150}{9} = \frac{155}{9}$$

$$9 \times \frac{155}{9} = 155$$.

Therefore, the answer is $$\boxed{155}$$.