1) \(x = 7 + 4\sqrt 3  = {\left( {2 + \sqrt 3 } \right)^2}\) (thỏa mãn điều kiện)

Suy ra \(\sqrt x  = 2 + \sqrt 3 .\)

\(A = \frac{{2 + \sqrt 3  - 1}}{{2 + \sqrt 3  - 2}} = \frac{{\sqrt 3  + 3}}{3}.\)

2) \(P = \left[ {\frac{{x - \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}} - \frac{x}{{\sqrt x \left( {\sqrt x  - 2} \right)}}} \right]:\frac{{\sqrt x  - 1}}{{\sqrt x  - 2}}\)

\(\begin{array}{l}
 = \frac{{x - \sqrt x  + 2 - \sqrt x \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}.\frac{{\sqrt x  - 2}}{{\sqrt x  - 1}} = \frac{{2 - 2\sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \frac{{2\left( {1 - \sqrt x } \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} =  - \frac{2}{{\sqrt x  + 1}}
\end{array}\)

Vậy \(P =  - \frac{2}{{\sqrt x  + 1}}\) với \(x > 0;x \ne 1;x \ne 4.\)

3) Ta có \(P\sqrt x  = \frac{{ - 2\sqrt x }}{{\sqrt x  + 1}} \ge  - \frac{3}{2} \Rightarrow \sqrt x  \le 3 \Rightarrow 0 \le x \le 9.\)

Mà \(x \in Z,x > 0,x \ne 1,x \ne 4 \Rightarrow x \in \left\{ {2;3;5;6;7;8;9} \right\}.\)