Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 4;x \ne 16\\
K = A.B = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 4}}\\
= \dfrac{{2\sqrt x + 8 - 9}}{{\sqrt x + 4}}\\
= \dfrac{{2\left( {\sqrt x + 4} \right) - 9}}{{\sqrt x + 4}}\\
= 2 - \dfrac{9}{{\sqrt x + 4}}\\
K \in Z\\
\Leftrightarrow \dfrac{9}{{\sqrt x + 4}} \in Z\\
\Leftrightarrow \sqrt x + 4 \in Ư\left( 9 \right)\\
\Leftrightarrow \sqrt x + 4 = 9\left( {do:\sqrt x + 4 > 4} \right)\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25\left( {tmdk} \right)\\
Vậy\,x = 25
\end{array}$
