Đáp án:
$\begin{array}{l}
A = \left( {\frac{1}{{\sqrt x + 3}} + \frac{{\sqrt x + 9}}{{x - 9}}} \right).\frac{{\sqrt x }}{2}\left( {x > 0;x \ne 9} \right)\\
a)A = \left( {\frac{1}{{\sqrt x + 3}} + \frac{{\sqrt x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right).\frac{{\sqrt x }}{2}\\
= \frac{{\sqrt x - 3 + \sqrt x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\frac{{\sqrt x }}{2}\\
= \frac{{2\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\frac{{\sqrt x }}{2}\\
= \frac{{2\left( {\sqrt x + 3} \right).\sqrt x }}{{2\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{2\sqrt x }}{{\sqrt x - 3}}\\
b)\,x = 16\left( {tmdk;x > 0;x \ne 9} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow A = \frac{{2\sqrt x }}{{\sqrt x - 3}} = \frac{{2.4}}{{4 - 3}} = 8\\
Vậy\,khi\,x = 16 \Rightarrow A = 8\\
c)\\
x > 0;x \ne 9\\
A = \frac{{2\sqrt x }}{{\sqrt x - 3}} = \frac{{2\sqrt x - 6 + 6}}{{\sqrt x - 3}}\\
= \frac{{2\left( {\sqrt x - 3} \right) + 6}}{{\sqrt x - 3}} = 2 + \frac{6}{{\sqrt x - 3}}\\
A \in Z\\
\Rightarrow \frac{6}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in Ư\left( 6 \right)\\
\Rightarrow \left( {\sqrt x - 3} \right) \in {\rm{\{ }} - 2; - 1;0;1;2;3;6\} \\
\left( {do:x > 0 \Rightarrow \sqrt x - 3 > - 3} \right)\\
\Rightarrow \sqrt x \in {\rm{\{ }}1;2;3;4;5;6;9\} \\
\Rightarrow x \in {\rm{\{ }}1;4;9;16;25;36;81\} \\
Mà:x \ne 9 \Rightarrow x \in {\rm{\{ }}1;4;16;25;36;81\}
\end{array}$
