Đáp án:

$\begin{array}{l}
a)x = 16\left( {tm} \right)\\
 \Rightarrow \sqrt x  = 4\\
 \Rightarrow A = \frac{{4 - 5}}{4} =  - \frac{1}{4}\\
b)B = \frac{{\sqrt x }}{{\sqrt x  - 5}} - \frac{{3\sqrt x }}{{x - 25}}\\
 = \frac{{\sqrt x \left( {\sqrt x  + 5} \right) - 3\sqrt x }}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}\\
 = \frac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}\\
P = A.B\\
 = \frac{{\sqrt x  - 5}}{{\sqrt x }}.\frac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}\\
 = \frac{{\sqrt x .\left( {\sqrt x  + 2} \right)}}{{\sqrt x .\left( {\sqrt x  + 5} \right)}}\\
 = \frac{{\sqrt x  + 2}}{{\sqrt x  + 5}}\\
c)P = \frac{{\sqrt x  + 2}}{{\sqrt x  + 5}} = 1 - \frac{3}{{\sqrt x  + 5}} < 1\\
 \Rightarrow P > {P^2}\\
2)a)5\sqrt {4x - 8}  - 3\sqrt {9x - 18}  = 4\left( {dkx \ge 2} \right)\\
 \Rightarrow 5.2\sqrt {x - 2}  - 3.3\sqrt {x - 2}  = 4\\
 \Rightarrow \sqrt {x - 2}  = 4\\
 \Rightarrow x - 2 = 16\\
 \Rightarrow x = 18\left( {tm} \right)\\
b)\sqrt {x - 1}  - \sqrt {x - 4}  = 1\left( {dk:x \ge 4} \right)\\
 \Rightarrow \sqrt {x - 1}  = \sqrt {x - 4}  + 1\\
 \Rightarrow x - 1 = x - 4 + 2\sqrt {x - 4}  + 1\\
 \Rightarrow 2\sqrt {x - 4}  = 2\\
 \Rightarrow \sqrt {x - 4}  = 1\\
 \Rightarrow x - 4 = 1\\
 \Rightarrow x = 5\left( {tm} \right)
\end{array}$