$\begin{array}{l}\quad A = \left(\dfrac{x+2}{\sqrt x + 1} - \sqrt x\right):\left(\dfrac{4- \sqrt x}{x-1} - \dfrac{\sqrt x}{\sqrt x+1}\right)\\ a)\quad ĐKXĐ: \begin{cases}x \geq 0\\x - 1 \ne 0\end{cases}\Leftrightarrow \begin{cases}x \geq 0\\x \ne 1\end{cases}\\ \quad A = \dfrac{x+2 - \sqrt x(\sqrt x + 1)}{\sqrt x + 1}:\left[\dfrac{4- \sqrt x}{(\sqrt x-1)(\sqrt x+1)} - \dfrac{\sqrt x(\sqrt x - 1)}{(\sqrt x-1)(\sqrt x+1)}\right]\\ \to A = \dfrac{x + 2 - x - \sqrt x}{\sqrt x + 1}:\dfrac{4 - \sqrt x -x + \sqrt x}{(\sqrt x-1)(\sqrt x+1)}\\ \to A = \dfrac{2 - \sqrt x}{\sqrt x+1}:\dfrac{4 - x}{(\sqrt x-1)(\sqrt x+1)}\\ \to A = \dfrac{2 - \sqrt x}{\sqrt x+1}\cdot\dfrac{(\sqrt x-1)(\sqrt x+1)}{(2-\sqrt x)(2 + \sqrt x)}\\ \to A = \dfrac{\sqrt x-1}{\sqrt x+2}\\ a)\quad A >0\\ \to \dfrac{\sqrt x - 1}{\sqrt x+2} >0\\ \text{Ta có:}\\ \quad \sqrt x \geq 0\\ \to \sqrt x + 2 >0\\ \text{Do đó:}\\ \quad \dfrac{\sqrt x - 1}{\sqrt x+2} >0\Leftrightarrow \sqrt x - 1 >0\\ \to \sqrt x > 1\\ \to x >1\\ c)\quad A = \dfrac{\sqrt x-1}{\sqrt x + 2}\\ \to A = \dfrac{\sqrt x + 2 - 3}{\sqrt x + 2}\\ \to A = 1 - \dfrac{3}{\sqrt x+2}\\ \text{Ta có:}\\ \quad \sqrt x \geq 0\\ \to \sqrt x + 2\geq 2\\ \to \dfrac{1}{\sqrt x +2} \leq \dfrac{1}{2}\\ \to -\dfrac{3}{\sqrt x+2} \geq -\dfrac32\\ \to 1 - \dfrac{3}{\sqrt x+2} \geq -\dfrac12\\ \to A \geq -\dfrac12\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \sqrt x = 0\Leftrightarrow x = 0\\ Vậy\,\,\min A = -\dfrac12 \Leftrightarrow x = 0 \end{array}$
