Đáp án:
$\begin{array}{l}
H = \dfrac{{2\left( {x + 4} \right)}}{{x - 3\sqrt x - 4}} + \dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{8}{{\sqrt x - 4}}\\
= \dfrac{{2x + 8 + \sqrt x \left( {\sqrt x - 4} \right) - 8\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{2x + 8 + x - 4\sqrt x - 8\sqrt x - 8}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{3x - 12\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 1}}\\
I = \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\dfrac{{2\sqrt x + 1}}{{x + \sqrt x - 2}}\\
= \dfrac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{2\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x + 1}}.\left( {\sqrt x + 2} \right)\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
K = \left( {\dfrac{{x - 7\sqrt x + 12}}{{x - 4\sqrt x + 3}} + \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{\sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{x - 7\sqrt x + 12 + \sqrt x - 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{x - 6\sqrt x + 9}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x - 3} \right)}^2}}}.\left( {\sqrt x + 3} \right)\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 1}}\\
M = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {5\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
N = \left( {\dfrac{{3x + \sqrt {9x} - 3}}{{x + \sqrt x - 2}} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 2}}} \right):\dfrac{1}{{x - 1}}\\
= \dfrac{{3x + 3\sqrt x - 3 + \sqrt x + 2 + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}.\left( {x - 1} \right)\\
= \dfrac{{3x + 5\sqrt x - 2}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\left( {3\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \left( {3\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
= 3x + 2\sqrt x - 1
\end{array}$
