Đáp án:

$1)\\ a)\text{ĐKXĐ:}  \left\{\begin{array}{l} x > 0\\x \ne 1 \\ x \ne 4\end{array} \right.\\ A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ b)x>4\\ 2)\\  a)Q=\dfrac{3}{\sqrt{x}+2}\\ b)x=\dfrac{1}{4}\\ c) x=1.$

Giải thích các bước giải:

$1)\\ A=\left(1-\dfrac{4}{\sqrt{x}+1}+\dfrac{1}{x-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\\ a)\text{ĐKXĐ: }\left\{\begin{array}{l} x \ge 0\\ \sqrt{x}+1 \ne 0 \\ x-1 \ne 0 \\ \dfrac{x-2\sqrt{x}}{x-1} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x} \ne -1 \\ x \ne 1 \\ x-2\sqrt{x}\ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\x \ne 1 \\ \sqrt{x}(\sqrt{x}-2)\ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\x \ne 1 \\ \sqrt{x}\ne 0 \\\sqrt{x}-2\ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\x \ne 1 \\ x \ne 0 \\ x \ne 4\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\x \ne 1 \\ x \ne 4\end{array} \right.\\ A=\left(1-\dfrac{4}{\sqrt{x}+1}+\dfrac{1}{x-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\\ =\left(1-\dfrac{4}{\sqrt{x}+1}+\dfrac{1}{(\sqrt{x}+1)(\sqrt{x}-1)}\right):\dfrac{x-2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\left(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{4(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{1}{(\sqrt{x}+1)(\sqrt{x}-1)}\right):\dfrac{x-2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)-4(\sqrt{x}-1)+1}{(\sqrt{x}+1)(\sqrt{x}-1)}.\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{x-2\sqrt{x}}\\ =\dfrac{x-4\sqrt{x}+4}{(\sqrt{x}+1)(\sqrt{x}-1)}.\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-2)}\\ =\dfrac{(\sqrt{x}-2)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}.\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-2)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ b)A>0\\ \Leftrightarrow \dfrac{\sqrt{x}-2}{\sqrt{x}}>0\\ \Leftrightarrow \sqrt{x}-2>0(\text{Do } \sqrt{x}>0 \ \forall x > 0;x \ne 1; x \ne 4)\\ \Leftrightarrow \sqrt{x}>2\\ \Leftrightarrow x>4\\ 2)\\ Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\\ a)\text{ĐKXĐ: } \left\{\begin{array}{l} x \ge 0 \\ 2+\sqrt{x} \ne 0 \\ 2-\sqrt{x} \ne 0 \\ x -4 \ne 0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x} \ne -2 \\ \sqrt{x} \ne 2 \\ x  \ne 4\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x  \ne 4\end{array} \right.\\ Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{(2+\sqrt{x})(2-\sqrt{x})}\\ =\dfrac{2(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}+\dfrac{2+\sqrt{x}}{(2-\sqrt{x})(2+\sqrt{x})}-\dfrac{2\sqrt{x}}{(2+\sqrt{x})(2-\sqrt{x})}\\ =\dfrac{2(2-\sqrt{x})+2+\sqrt{x}-2\sqrt{x}}{(2+\sqrt{x})(2-\sqrt{x})}\\ =\dfrac{6-3\sqrt{x}}{(2+\sqrt{x})(2-\sqrt{x})}\\ =\dfrac{3(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}\\ =\dfrac{3}{\sqrt{x}+2}\\ b)Q=\dfrac{6}{5}\\ \Leftrightarrow Q-\dfrac{6}{5}=0\\ \Leftrightarrow \dfrac{3}{\sqrt{x}+2}-\dfrac{6}{5}=0\\ \Leftrightarrow \dfrac{15-6(\sqrt{x}+2)}{5(\sqrt{x}+2)}=0\\ \Leftrightarrow \dfrac{3-6\sqrt{x}}{5(\sqrt{x}+2)}=0\\ \Leftrightarrow 3-6\sqrt{x}=0\\ \Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{1}{4}\\ c)Q \in \mathbb{Z}\\ \Leftrightarrow \dfrac{3}{\sqrt{x}+2}\in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}+2) \in Ư(3)\\ \Leftrightarrow (\sqrt{x}+2) \in \{\pm 1; \pm 3\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}+2=-3 \\ \sqrt{x}+2=-1 \\ \sqrt{x}+2 =1\\ \sqrt{x}+2=3\end{array} \right.\Leftrightarrow \left[\begin{array}{l} \sqrt{x}=-5 \\ \sqrt{x}=-3 \\ \sqrt{x} =-1\\ \sqrt{x}=1\end{array} \right.\Leftrightarrow x=1.$