(1,5 điểm) Cho hai biểu thức:

1) Thay x = 25 (tmđk) vào A ta được \(A = \frac{{\sqrt {25}  - 1}}{{\sqrt {25} }} = \frac{4}{5}\)

Vậy với \(x = 25\) thì \(A = \frac{4}{5}\)

\(2)B = \frac{{2\sqrt x }}{{\sqrt x  + 3}} + \frac{4}{{\sqrt x  - 3}} - \frac{{x - 4\sqrt x  + 15}}{{x - 9}}\)\[ = \frac{{2\sqrt x \left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} + \frac{{4\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} - \frac{{x - 4\sqrt x  + 15}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\]

\( = \frac{{2x - 6\sqrt x  + \left( {4\sqrt x  + 12} \right) - \left( {x - 4\sqrt x  + 15} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\)\( = \frac{{x + 2\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\)

\( = \frac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} = \frac{{\sqrt x  - 1}}{{\sqrt x  - 3}}\)

3) \(P = A:B = \frac{{\sqrt x  - 1}}{{\sqrt x }}:\frac{{\sqrt x  - 1}}{{\sqrt x  - 3}} = \frac{{\sqrt x  - 3}}{{\sqrt x }}\) (đkbs: \(x \ne 1\))

\[\left| P \right| + P = 0 \Rightarrow \left| P \right| =  - P \Rightarrow P \le 0 \Rightarrow \frac{{\sqrt x  - 3}}{{\sqrt x }} \le 0 \Leftrightarrow x \le 9\]

KHĐK  \( \Rightarrow 0 < x < 9;x \ne 1\)

Mà \(x \in \mathbb{Z} \Rightarrow x \in \left\{ {2;3;4;5;6;7;8} \right\}\)