Đáp án:
$\begin{array}{l}
b)\dfrac{{x - 12\sqrt x + 6}}{{x - 36}} + 2\sqrt x \\
= \dfrac{{x - 12\sqrt x + 6 + 2\sqrt x \left( {x - 36} \right)}}{{x - 36}}\\
= \dfrac{{x - 12\sqrt x + 6 + 2x\sqrt x - 72\sqrt x }}{{x - 36}}\\
= \dfrac{{2x\sqrt x + x - 84\sqrt x + 6}}{{x - 36}}\\
c)\dfrac{{x - 3\sqrt x + 2}}{{x - \sqrt x }} + \sqrt x - 1\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \left( {\sqrt x - 1} \right)\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x }} + \left( {\sqrt x - 1} \right)\\
= \dfrac{{\sqrt x - 2 + \sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{\sqrt x - 2 + x - \sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - 2}}{{\sqrt x }}\\
d)\dfrac{{x + \sqrt x - 6}}{{x - 9}} + 2\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} + 2\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} + 2\\
= \dfrac{{\sqrt x - 2 + 2\sqrt x - 6}}{{\sqrt x - 3}}\\
= \dfrac{{3\sqrt x - 8}}{{\sqrt x - 3}}\\
e)\sqrt x + \dfrac{{x + 5\sqrt x + 6}}{{x - 4}}\\
= \sqrt x + \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \sqrt x + \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) + \sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{x - 2\sqrt x + \sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{x - \sqrt x + 3}}{{\sqrt x - 2}}
\end{array}$
