Đáp án:
$\begin{array}{l}
a)A = \left( {\dfrac{{x - 3\sqrt x }}{{x - 9}} - 1} \right)\\
:\left( {\dfrac{{9 - x}}{{x + \sqrt x - 6}} + \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} - 1} \right)\\
:\dfrac{{9 - x + \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x + 3}} - 1} \right).\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{9 - x + x - 9 - {{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{{\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{ - {{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{{ - 3}}{1}.\dfrac{1}{{ - \left( {\sqrt x - 2} \right)}}\\
= \dfrac{3}{{\sqrt x - 2}}\\
d)A \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 2}} \in Z\\
+ Khi:\dfrac{3}{{\sqrt x - 2}} = - 1\\
\Leftrightarrow \sqrt x - 2 = - 3\\
\Leftrightarrow \sqrt x = - 1\left( {ktm} \right)\\
+ Khi:\dfrac{3}{{\sqrt x - 2}} = - 3\\
\Leftrightarrow \sqrt x - 2 = - 1\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tm} \right)
\end{array}$
Vậy giá trị nguyên âm lớn nhất của A là -3 khi x=1
$\begin{array}{l}
e)A \ge 0\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 2}} \ge 0\\
\Leftrightarrow \sqrt x - 2 > 0\\
\Leftrightarrow x > 4\\
B = \left( {x + 5} \right).A\\
= \left( {x + 5} \right).\dfrac{3}{{\sqrt x - 2}}\\
= \dfrac{{3x + 15}}{{\sqrt x - 2}}\\
= \dfrac{{3x - 12 + 27}}{{\sqrt x - 2}}\\
= \dfrac{{3.\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) + 27}}{{\sqrt x - 2}}\\
= 3.\left( {\sqrt x + 2} \right) + \dfrac{{27}}{{\sqrt x - 2}}\\
= 3\left( {\sqrt x - 2} \right) + \dfrac{{27}}{{\sqrt x - 2}} + 12\\
Theo\,Co - si:\\
3\left( {\sqrt x - 2} \right) + \dfrac{{27}}{{\left( {\sqrt x - 2} \right)}} \ge 2.\sqrt {3\left( {\sqrt x - 2} \right).\dfrac{{27}}{{\left( {\sqrt x - 2} \right)}}} = 2.9 = 18\\
\Leftrightarrow B \ge 18 + 12\\
\Leftrightarrow B \ge 30\\
\Leftrightarrow GTNN:\left( {x + 5} \right).A = 30\,\\
Khi:3\left( {\sqrt x - 2} \right) = \dfrac{{27}}{{\sqrt x - 2}} \Leftrightarrow x = 25
\end{array}$
