$\begin{array}{l}1b) \, A = \dfrac{x}{\sqrt x + 2}\\ E = \dfrac{2A}{\sqrt x} = \dfrac{\dfrac{2x}{\sqrt x + 2}}{\sqrt x} = \dfrac{2x}{x + 2\sqrt x} = \dfrac{2}{1 + \dfrac{2}{\sqrt x}}\\ E \in \Bbb Z \Leftrightarrow \dfrac{2}{1 + \dfrac{2}{\sqrt x}} \in \Bbb Z\\ \Leftrightarrow 1 + \dfrac{2}{\sqrt x} \in Ư(2) = \left\{-2;-1;1;2\right\}\\ Do\,\,\sqrt x > 0\\ nên\,\, 1 + \dfrac{2}{\sqrt x} > 0\\ \Rightarrow 1 + \dfrac{2}{\sqrt x} = \left\{1;2\right\}\\ \star \,\,1 + \dfrac{2}{\sqrt x} = 1 \Leftrightarrow \dfrac{2}{\sqrt x} = 0 \quad \text{(vô nghiệm)}\\ \star\,\, 1 + \dfrac{2}{\sqrt x} = 2 \Leftrightarrow \dfrac{2}{\sqrt x} = 1 \Leftrightarrow x = 4 \quad \text{(Không thỏa ĐKXĐ)}\\ \text{Vậy không có x thỏa yêu cầu bài toán}\\ 3)\, P = \left(\dfrac{\sqrt x - 3}{2 - \sqrt x} + \dfrac{\sqrt x + 2}{3 + \sqrt x} - \dfrac{9 - x}{x + \sqrt x - 6}\right):\left(1 - \dfrac{3\sqrt x - 9}{x - 9}\right)\\ =\left[\dfrac{\sqrt x - 3}{2 - \sqrt x} + \dfrac{\sqrt x + 2}{3 + \sqrt x} - \dfrac{x-9}{(2 - \sqrt x)(3 + \sqrt x)}\right]:\left[1 - \dfrac{3(\sqrt x - 3)}{(\sqrt x - 3)(\sqrt x + 3)}\right]\\ =\left[\dfrac{(\sqrt x - 3)(3 + \sqrt x)}{(2 - \sqrt x)(3 + \sqrt x)} + \dfrac{(\sqrt x + 2)(2 - \sqrt x)}{(3 + \sqrt x)(2 - \sqrt x)} - \dfrac{x - 9}{(2 - \sqrt x)(3 + \sqrt x)}\right]:\left(1 - \dfrac{3}{\sqrt x + 3}\right)\\ =\left[\dfrac{x - 9 + 4 -x - x + 9}{(2 - \sqrt x)(3 + \sqrt x)}\right]: \left(\dfrac{\sqrt x + 3 - 3}{\sqrt x + 3} \right)\\ = \dfrac{4 - x}{(2 - \sqrt x)(3 + \sqrt x)}\cdot \dfrac{\sqrt x + 3}{\sqrt x}\\ = \dfrac{(2- \sqrt x)(2 + \sqrt x)}{(2 - \sqrt x)(3 + \sqrt x)}\cdot \dfrac{\sqrt x + 3}{\sqrt x}\\ = \dfrac{\sqrt x + 2}{\sqrt x}\\ Ta\,\,có:\\x = \dfrac{\sqrt{4 + 2\sqrt3}.(\sqrt3 - 1)}{\sqrt{6 + 2\sqrt5} - \sqrt5}\\ = \dfrac{(\sqrt3 + 1)(\sqrt 3 - 1)}{(\sqrt5 + 1) - \sqrt5}\\ = 2\\ \text{Thay vào P ta được:}\\ P = \dfrac{\sqrt2 + 2}{\sqrt 2} = 1 + \sqrt2\end{array}$
