$\begin{array}{l}P = \left(\dfrac{\sqrt x + 1}{\sqrt x -1} +\dfrac{\sqrt x}{\sqrt x + 1} + \dfrac{\sqrt x}{1 - \sqrt x}\right): \left(\dfrac{\sqrt x + 1}{\sqrt x -1} + \dfrac{1 -\sqrt x}{\sqrt x + 1}\right)\\ ĐK: \, x \geq 0; \, x \ne 1\\ P = \dfrac{(\sqrt x + 1)^2 + \sqrt x(\sqrt x -1)-\sqrt x(\sqrt x + 1)}{(\sqrt x - 1)(\sqrt x + 1)}:\dfrac{(\sqrt x +1)^2 - (\sqrt x -1)^2}{(\sqrt x - 1)(\sqrt x + 1)}\\ = \dfrac{x + 2\sqrt x + 1 + x - \sqrt x- x - \sqrt x}{(\sqrt x - 1)(\sqrt x + 1)}:\dfrac{(\sqrt x+1 +\sqrt x - 1)(\sqrt x + 1 - \sqrt x + 1)}{(\sqrt x - 1)(\sqrt x + 1)}\\ =\dfrac{x +1}{(\sqrt x - 1)(\sqrt x + 1)}\cdot \dfrac{(\sqrt x - 1)(\sqrt x + 1)}{4\sqrt x}\\ = \dfrac{x +1}{4\sqrt x}\\ Xét\,\,P - \dfrac{1}{2}\\ = \dfrac{x +1}{4\sqrt x} - \dfrac{1}{2}\\ = \dfrac{x -2\sqrt x + 1}{4\sqrt x}\\ = \dfrac{(\sqrt x - 1)^2}{4\sqrt x} \geq 0\\\text{Do P xác định}\\\Rightarrow \sqrt x > 0\\\text{Ta lại có: x $\ne$ 1}\\\Rightarrow \dfrac{(\sqrt x - 1)^2}{4\sqrt x} >0\\\text{Vậy $P > \dfrac{1}{2}$}\end{array}$
