Đáp án:

1) \(Max = 1\)

Giải thích các bước giải:

\(\begin{array}{l}
1)P = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
Do:x \ge 0\\
 \to \sqrt x  \ge 0\\
 \to \sqrt x  + 3 \ge 3\\
 \to \dfrac{3}{{\sqrt x  + 3}} \le \dfrac{3}{3}\\
 \to \dfrac{3}{{\sqrt x  + 3}} \le 1\\
 \to Max = 1\\
 \Leftrightarrow x = 0\\
2)A = \left[ {\dfrac{{\sqrt x  - 2 + \sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}} \right].\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
 = \dfrac{{2\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
 = \dfrac{2}{{\sqrt x  + 2}}\\
\dfrac{{7A}}{3} = \dfrac{{7.2}}{{3\left( {\sqrt x  + 2} \right)}} = \dfrac{{14}}{{3\left( {\sqrt x  + 2} \right)}}\\
\dfrac{{7A}}{3} \in Z\\
 \to \dfrac{{14}}{{3\left( {\sqrt x  + 2} \right)}} \in Z\\
 \Leftrightarrow 3\left( {\sqrt x  + 2} \right) \in U\left( {14} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3\left( {\sqrt x  + 2} \right) = 14\\
3\left( {\sqrt x  + 2} \right) =  - 14\\
3\left( {\sqrt x  + 2} \right) = 7\\
3\left( {\sqrt x  + 2} \right) =  - 7\\
3\left( {\sqrt x  + 2} \right) = 2\\
3\left( {\sqrt x  + 2} \right) =  - 2\\
3\left( {\sqrt x  + 2} \right) = 1\\
3\left( {\sqrt x  + 2} \right) =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x  + 2 = \dfrac{{14}}{3}\left( l \right)\\
\sqrt x  + 2 =  - \dfrac{{14}}{3}\left( l \right)\\
\sqrt x  + 2 = \dfrac{7}{3}\left( l \right)\\
\sqrt x  + 2 =  - \dfrac{7}{3}\left( l \right)\\
\sqrt x  + 2 = \dfrac{2}{3}\left( l \right)\\
\sqrt x  + 2 =  - \dfrac{2}{3}\left( l \right)\\
\sqrt x  + 2 = \dfrac{1}{3}\left( l \right)\\
\sqrt x  + 2 =  - \dfrac{1}{3}\left( l \right)
\end{array} \right.
\end{array}\)

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