$\left(\sqrt{4+\frac{1}{x}}-2 \right) \cdot \left(\sqrt{4+\frac{1}{x}}+2\right)$
I get $\large\frac{1}{x}$ because the square roots go away and the $2$s multiply to make $-4$, so it's:
$4 + \large\frac{1}{x} - 4 = \large\frac{1}{x}$
Is this right? Or no?
Yes. You are right.
Using $x^2-y^2=(x-y)(x+y)$, so you will have
$$(\sqrt{4+\frac 1x}{}-2)\times( \sqrt{4+\frac 1x}{}+2)=4+\frac1x-2^2=\frac1x$$