ĐK: $x\ge 0,x\ne 1$
$\dfrac{x+\sqrt x+1}{\sqrt x-1}\\=\dfrac{x-\sqrt x+2\sqrt x+1}{\sqrt x-1}\\=\dfrac{\sqrt x(\sqrt x-1)+2\sqrt x-2+3}{\sqrt x-1}\\=\sqrt x+\dfrac{2(\sqrt x-1)+3}{\sqrt x-1}\\=\sqrt x+2+\dfrac{3}{\sqrt x-1}$
Vì $\dfrac{x+\sqrt x+1}{\sqrt x-1}∈\Bbb Z$
$→\sqrt x +2+\dfrac{3}{\sqrt x-1}∈\Bbb Z$
$→\begin{cases}\sqrt x ∈\Bbb Z\\3\vdots \sqrt x-1\end{cases}$
$·3\vdots x-1\\→\sqrt x-1∈Ư(3)=\{±1;±3\}$
Ta có bảng:
$\begin{array}{|c|c|c|}\hline \sqrt x -1&1&-1&3&-3\\\hline \sqrt x &2&0&4&-2\\\hline\end{array}$
Vì $\sqrt x -2≥-2$
$→\sqrt x ∈\{2;0;4\}$ (thỏa mãn điều kiện $x∈\Bbb Z$)
$↔x∈\{4;0;16\}$
mà $x\ge 0,x\ne 1$
$⇒x∈\{4;0;16\}$
Vậy $x∈\{4;0;16\}$
