Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1;x \ne 4\\
B = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\
= \dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)Dkxd:x > 0;x \ne 1;x \ne 4\\
B = \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 - 1}}{{\sqrt x - 2}} = 1 - \dfrac{1}{{\sqrt x - 2}}\\
B \in Z\\
\Rightarrow \left( {\sqrt x - 2} \right) \in U\left( 1 \right) = \left\{ { - 1;1} \right\}\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {ktm} \right)\\
x = 9\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 9
\end{array}$
