Đáp án:
$Dkxd:x \ge 0;x \ne 1$
$\begin{array}{l}
K = \dfrac{{x + \sqrt x + 7}}{{\sqrt x - 1}}\\
= \dfrac{{x - \sqrt x + 2\sqrt x - 2 + 9}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 2\left( {\sqrt x - 1} \right) + 9}}{{\sqrt x - 1}}\\
= \sqrt x + 2 + \dfrac{9}{{\sqrt x - 1}}\\
K \in Z\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
\dfrac{9}{{\sqrt x - 1}} \in Z
\end{array} \right.\\
\Rightarrow \left( {\sqrt x - 1} \right) \in \left\{ { - 1;1;3;9} \right\}\left( {do:\sqrt x - 1 \ge - 1} \right)\\
\Rightarrow \sqrt x \in \left\{ {0;2;4;10} \right\}\\
\Rightarrow x \in \left\{ {0;4;16;100} \right\}
\end{array}$
