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 $(4)$?

As you have $\frac{1}{\sqrt n}$ in the numerator, first multiply numerator and denominator by $\sqrt n$ and check if it gets simplified.
$$\frac{2\sqrt n +\frac1{\sqrt n} -3}{2\sqrt n -1}=\frac{2n + 1 -3\sqrt n}{\sqrt n(2\sqrt n - 1)}$$

If you let $\sqrt n = x$, the numerator becomes, $2x^2 - 3x +1 = 2x^2-2x-x+1 = (2x-1)(x-1) = \boxed{(2\sqrt n-1)(\sqrt n -1)}$

So,
$$\frac{2n + 1 -3\sqrt n}{\sqrt n(2\sqrt n - 1)} = \frac{(2\sqrt n-1)(\sqrt n -1)}{\sqrt n(2\sqrt n - 1)}=\frac{\sqrt n-1}{\sqrt n} = \frac{n-\sqrt n}{n}$$
