<s>[INST] Simplifying $\frac{2\sqrt n + \frac{1}{\sqrt n}-3}{2\sqrt n -1}$ Consider the term $$\dfrac{2\sqrt n + \dfrac{1}{\sqrt n}-3}{2\sqrt n -1} \tag1$$
I simplified it to
$$\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1} \tag2$$
or also
$$\dfrac{2 n+1-3 \sqrt n}{2 n-\sqrt{n}} \tag3$$
but apparently one can simplify this to
$$\dfrac{n-\sqrt n}{n} \tag4$$
but I don't really get to this expression.
How can one arrive at $(4)$?
 [/INST] You were almost there:

I simplified it to $\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1}$

From there, the key is to spot the $4n-1$ in the numerator and pull it out:
$$\begin{align}
\ldots &= \frac{4n-1}{4n-1} + \frac{-4\sqrt n  + \frac1{\sqrt n}}{4n-1}\\
&= 1 + \frac1{\sqrt n}\frac{-4n+1}{4n-1}\\
&= 1 - \frac1{\sqrt n}
\end{align}$$
</s>