Đáp án:
$\begin{array}{l}
B = \dfrac{{x + \sqrt x }}{{1 - x}} + \dfrac{{{{\left( {\sqrt x - 2} \right)}^2} - \sqrt x - x}}{{1 - \sqrt x }}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{x - 4\sqrt x + 4 - \sqrt x - x}}{{1 - \sqrt x }}\\
= \dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{{ - 5\sqrt x + 4}}{{1 - \sqrt x }}\\
= \dfrac{{ - 4\sqrt x + 4}}{{1 - \sqrt x }}\\
= \dfrac{{4\left( {1 - \sqrt x } \right)}}{{1 - \sqrt x }}\\
= 4\\
B = \left( {\dfrac{{\sqrt x }}{{2 + \sqrt x }} + \dfrac{{x + 4}}{{4 - x}}} \right):\dfrac{x}{{x - 2\sqrt x }}\\
= \dfrac{{\sqrt x \left( {2 - \sqrt x } \right) + x + 4}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{x}\\
= \dfrac{{2\sqrt x - x + x + 4}}{{2 + \sqrt x }}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{2\left( {\sqrt x + 2} \right)}}{{2 + \sqrt x }}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{ - 2}}{{\sqrt x }}
\end{array}$
