If  $\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2},x\ne 0,x\ne 3\text{ then x =}$

Let  $\sqrt{\frac{x}{x-3}}=t$

The given equation becomes 

$\begin{array}{rcl}t+\frac{1}{t}& =& \frac{5}{2}\\ & =& 2+\frac{1}{2}\end{array}$

If $t=2$ $\Rightarrow$ $t=\frac{1}{2}$ or 

If  $t=\frac{1}{2}\Rightarrow \frac{1}{t}= 2$

Hence, $\sqrt{\frac{x}{x-3}}=2$

Squaring on both sides 

$\begin{array}{rcl}\frac{x}{x-3}& =& 4\\ x& =& 4x-12\\ 3x& =& 12\\ x& =& 4\end{array}$

or 

$\sqrt{\frac{x}{x-3}}=\frac{1}{2}$

Squaring on both sides

$\begin{array}{rcl}\frac{x}{x-3}& =& \frac{1}{4}\\ 4x& =& x-3\\ 3x& =& -3\\ x& =& -1\end{array}$

Therefore, the roots are $\boxed{\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{4}}$