Đáp án:

$1\\a)x \in \{0;1;9;16\}\\ b) x \in \{0;1;9\}\\ c) x \in \{4;16;64\}\\ 2)\\ a)x \in \{1\}\\ b)x \in \{4\}\\ c)x \in \{0;1\}$

Giải thích các bước giải:

$1\\a)A=\dfrac{2}{\sqrt{x}-2}(x \ge 0; x \ne 4)\\ A=\dfrac{2}{\sqrt{x}-2} \in \mathbb{Z}; x \in \mathbb{Z}\\ \Rightarrow (\sqrt{x}-2) \in Ư(2)\\ \Leftrightarrow (\sqrt{x}-2) \in \{\pm 1; \pm 2\}\\ \Rightarrow x \in \{0;1;9;16\}\\ b)B=\dfrac{4}{\sqrt{x}+1}(x \ge 0)\\ B=\dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}; x \in \mathbb{Z}\\ \Rightarrow (\sqrt{x}+1) \in Ư(4)\\ \Leftrightarrow (\sqrt{x}+1) \in \{\pm 1; \pm 2; \pm 4\}\\ \Rightarrow x \in \{0;1;9\}\\ c)C=\dfrac{5}{\sqrt{x}-3}(x \ge 0; x \ne 9)\\ C=\dfrac{5}{\sqrt{x}-3} \in \mathbb{Z}; x \in \mathbb{Z}\\ \Rightarrow (\sqrt{x}-3) \in Ư(5)\\ \Leftrightarrow (\sqrt{x}-3) \in \{\pm 1; \pm 5\}\\ \Rightarrow x \in \{4;16;64\}\\ 2)\\ a)A=\dfrac{\sqrt{x}+1}{\sqrt{x}} (x>0)\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}} \in \mathbb{Z}\\ =1+\dfrac{1}{\sqrt{x}} \in \mathbb{Z}\\ \Rightarrow \dfrac{1}{\sqrt{x}} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow \sqrt{x} \in Ư(1)\\ \Leftrightarrow \sqrt{x} \in \{\pm 1\}\\ \Rightarrow x \in \{1\}\\ b)B=\dfrac{\sqrt{x}-2}{\sqrt{x}+3} (x\ge 0)\\ B=\dfrac{\sqrt{x}-2}{\sqrt{x}+3} \in \mathbb{Z}\\ =\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3} \in \mathbb{Z}\\ =1-\dfrac{5}{\sqrt{x}+3} \in \mathbb{Z}\\ \Rightarrow \dfrac{5}{\sqrt{x}+3} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}+3) \in Ư(5)\\ \Leftrightarrow (\sqrt{x}+3) \in \{\pm 1; \pm 5\}\\ \Rightarrow x \in \{4\}\\ c)C=\dfrac{1-\sqrt{x}}{\sqrt{x}+1} (x\ge 0)\\ C=\dfrac{1-\sqrt{x}}{\sqrt{x}+1} \in \mathbb{Z}\\ =\dfrac{-\sqrt{x}-1+2}{\sqrt{x}+1} \in \mathbb{Z}\\ =-1+\dfrac{2}{\sqrt{x}+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{2}{\sqrt{x}+1} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}+1) \in Ư(2)\\ \Leftrightarrow (\sqrt{x}+1) \in \{\pm 1; \pm 2\}\\ \Rightarrow x \in \{0;1\}$