\(\begin{array}{l}
A = \dfrac{3 -\sqrt x}{\sqrt x + 1}\\
B = \dfrac{\sqrt x - 1}{3 - \sqrt x } - \dfrac{9\sqrt x  +5}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)} - \dfrac{\sqrt x }{\sqrt x  +1}\qquad (x \geqslant 0;x \ne 9)\\
a)\\
\quad A = 4\\
\to \dfrac{3 - \sqrt x}{\sqrt x + 1}=4\\
\to 3 - \sqrt x = 4\left(\sqrt x + 1\right)\\
\to 5\sqrt x = -1\\
\to \sqrt x =- \dfrac15\quad \text{(vô lí)}\\
\text{Vậy không có giá trị x thỏa $A = 4$}\\
b)\\
\quad B = \dfrac{\left(1 - \sqrt x\right)\left(\sqrt x  +1\right)}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}-\dfrac{9\sqrt x+5}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)} - \dfrac{\sqrt x \left(\sqrt x -3\right)}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}\\
\to B = \dfrac{1 - x - 9\sqrt x - 5 - x + 3\sqrt x}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}\\
\to B = \dfrac{-2\left(x + 3\sqrt x + 2\right)}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}\\
\to B =\dfrac{-2\left(\sqrt x +1\right)\left(\sqrt x + 2\right)}{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}\\
\to B =\dfrac{-2\left(\sqrt x + 2\right)}{\sqrt x -3}\\
c)\\
\quad A = \dfrac{3-\sqrt x}{\sqrt x + 1}\\
\to A = -1 + \dfrac{4}{\sqrt x + 1}\\
\text{Ta có:}\\
\sqrt x \geqslant 0\quad \forall x\geqslant 0\\
\to \sqrt x  +1\geqslant 1\\
\to \dfrac{1}{\sqrt x + 1}\leqslant 1\\
\to \dfrac{4}{\sqrt x+1}\leqslant 4\\
\to -1 + \dfrac{4}{\sqrt x+ 1}\leqslant 3\\
\text{Hay}\ A \leqslant 3\\
\text{Dấu = xảy ra khi và chỉ khi}\ \sqrt x = 0\Leftrightarrow x = 0\\
\text{Vậy}\ \max A = 3\Leftrightarrow x = 0\\
d)\\
\quad P = AB\\
\to P = \dfrac{3-\sqrt x}{\sqrt x + 1}\cdot \dfrac{-2\left(\sqrt x + 2\right)}{\sqrt x - 3}\\
\to P = \dfrac{2\left(\sqrt x + 2\right)}{\sqrt x  +1}\\
\to P = 2  + \dfrac{2}{\sqrt x  +1}\\
P\in \Bbb Z \Leftrightarrow \dfrac{2}{\sqrt x + 1}\in \Bbb Z\\
\Leftrightarrow \sqrt x + 1 = \dfrac{2}{k}\quad (k\in\Bbb N^*)\\
\Leftrightarrow \sqrt x = \dfrac{2}{k} - 1\\
\text{Ta lại có:}\\
\sqrt x \geqslant 0\quad \forall x\geqslant 0\\
\Leftrightarrow \dfrac{2}{k} - 1\geqslant 0\\
\Leftrightarrow \dfrac{2-k}{k}\geqslant 0\\
\Leftrightarrow 0<k\leqslant 2\\
\Leftrightarrow k \in \{1;2\}\quad (Do\ k\in\Bbb N^*)\\
\text{Ta được:}\\
\begin{cases}\sqrt x = \dfrac21 - 1\\\sqrt x = \dfrac22 -1\end{cases}\\
\to \begin{cases}\sqrt x = 1\\\sqrt x = 0\end{cases}\\
\to \begin{cases}x = 1\\x  =0\end{cases}\\
\text{Vậy}\ x\in\{0;1\}
\end{array}\)