$Đk:x\ge0;x\ne4\\ P=A.B\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}-2}.\bigg(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}}{4-x}-\dfrac{2}{\sqrt{x}+2}\bigg)\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}-2}.\dfrac{\sqrt{x}(\sqrt{x}+2)-4\sqrt{x}-2(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}-2}.\dfrac{x-4\sqrt{x}+4}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{\sqrt{x}-5}{\sqrt{x}-2}.\dfrac{(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+2-7}{\sqrt{x}+2}\\ =1-\dfrac{7}{\sqrt{x}+2}$
Để $P<P^2$ thì $\left[\begin{matrix}P<0\\P>1\end{matrix}\right.$
$\Leftrightarrow \left[ \begin{array}{l}1-\dfrac{7}{\sqrt{x}+2}<0\\1-\dfrac{7}{\sqrt{x}+2}>1\end{array} \right.\\ \Leftrightarrow \left[\begin{matrix}\dfrac{7}{\sqrt{x}+2}>1\Leftrightarrow 0\le x<25\\\dfrac{7}{\sqrt{x}+2}<0\ (vô\ nghiệm)\end{matrix}\right.$
Kết hợp với điều kiện đề bài thì $0\le x<25; x\ne4$ là giá trị cần tìm.
