If $x$ is irrational, is $2x - \frac{1}{x}$ irrational?

Suppose $2x - \dfrac 1 x = a$ and $a$ is rational. Multiplying both sides by $x,$ we get $$ 2x^2 - 1 = ax. $$ This is a quadratic equation whose solution is $$ x = \frac{a\pm \sqrt{a^2 + 8}} 4 $$ Choose $a$ so that $\sqrt{a^2+8}$ is irrational, and then you have $x$ irrational and $2x-\dfrac 1 x$ rational.

Consider a rational number $q$ such that $\sqrt{q^2+8}\notin\mathbb Q$ – not a very hard condition to fulfill (any integral $q>1$ makes this happen, for example). Then, $x=\frac{q\pm\sqrt{q^2+8}}{4}$, for both choices of the sign, will be irrational, while $2x-\frac 1x=q$ will be rational. In this manner, we can generate infinitely many counterexamples.