Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 4\\
B = \dfrac{{2\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{2\sqrt x + 2 - 2}}{{\sqrt x + 1}}\\
= 2 - \dfrac{2}{{\sqrt x + 1}}\\
B \in Z\\
\Rightarrow \dfrac{2}{{\sqrt x + 1}} \in Z\\
Do:\sqrt x > 0;\sqrt x \ne 2\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 0\left( {ktm} \right)\\
\sqrt x = 1
\end{array} \right.\\
\Rightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
