a)
$\begin{array}{l} B = \dfrac{{3x - 4}}{{x - 2\sqrt x }} - \dfrac{{\sqrt x + 2}}{{\sqrt x }} + \dfrac{{\sqrt x - 1}}{{2 - \sqrt x }}\\ B = \dfrac{{3x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \dfrac{{\sqrt x + 2}}{{\sqrt x }} - \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}}\\ B = \dfrac{{3x - 4 - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\ B = \dfrac{{3x - 4 - \left( {x - 4} \right) - \left( {x - \sqrt x } \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\ B = \dfrac{{x + \sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}} = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x - 2}} = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} = C\\ b)B = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} > \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{1}{2} > 0\\ \Leftrightarrow \dfrac{{2\sqrt x + 2 - \left( {\sqrt x - 2} \right)}}{{2\left( {\sqrt x - 2} \right)}} > 0\\ \Leftrightarrow \dfrac{{\sqrt x + 4}}{{2\left( {\sqrt x - 2} \right)}} > 0\\ \Leftrightarrow 2\left( {\sqrt x - 2} \right) > 0\\ \Leftrightarrow \sqrt x > 2 \Leftrightarrow x > 4 \Rightarrow \min x\in \mathbb{Z} = 5 \end{array}$
