If $x - \sqrt{ \frac{2}{x} } = 3 $ then $x-\sqrt{2x} = ?$ Please find the value without calculate the value of $x$. I tried to multipy it by $x$, I tried to square it but I still can't find the solution. I tried to make some equation that my help to solve this:
$$x^{2} + \frac{2}{x}-2\sqrt{2x} =9, $$
$$x^{2} - \sqrt{2x} = 3x,$$
$$x^2 - \frac{2}{x} = 3\left(x+\sqrt{\frac{2}{x}}\right).$$

\begin{align}
&x - \sqrt{\frac{2}{x}} = 3\\
&x\sqrt{x} - \sqrt{2} = 3\sqrt{x}\\
&\sqrt{x}(x-3)=\sqrt{2}\\
&x(x-3)^2=2\\
&(x-2)(x^2-4x+1)=0
\end{align}
Note that $x=2$ is not a solution, so
$$x^2-4x+1=0 \implies x=2+\sqrt{3} \text{ or } x=2-\sqrt{3}$$
The second is not a solution either, so $x= 2+ \sqrt{3}.$
In that case, $x - \sqrt{2x} = 2+ \sqrt{3} - \sqrt{2(2 + \sqrt{3})} = 2 + \sqrt{3} - \sqrt{(\sqrt{3}+1)^2} = 1.$
