$\dfrac{4\sqrt{x}+1}{2\sqrt{x}-1}$
$=\dfrac{4\sqrt{x}-2+3}{2\sqrt{x}-1}$
$=2+\dfrac{3}{2\sqrt{x}-1}$
Để $Q∈\mathbb{Z}$ thì $3\vdots 2\sqrt{x}-1$
$→2\sqrt{x}-1∈Ư(3)=\{1;-1;3;-3\}$
Ta có bảng giá trị
\begin{array}{|l|r|} \hline 2\sqrt x - 1 & 1&-1&3 &-3\\ \hline \quad\sqrt x&1&0&4&-2\\ \hline \,\,\quad x&1&0&16&ktm\\ \hline \end{array}
$→x=\{1;0;16\}$
Vậy $x=\{1;0;16\}$ thì $Q∈ \mathbb{Z}$
