`~rai~`
\(\text{a)Với x}\ne 4;x\ge 0,\text{ta có:}\\A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\\quad=\dfrac{x}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\\quad=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\\quad=\dfrac{x+2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\\\quad=\dfrac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\\\quad=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\\\text{Vậy với x}\ne 4;x\ge 0\text{ thì A=}\dfrac{\sqrt{x}}{\sqrt{x}-2}.\\\text{b)Thay x=25(thỏa mãn) vào biểu thức A được:}\\A=\dfrac{\sqrt{25}}{\sqrt{25}-2}=\dfrac{5}{3}.\\\text{Vậy giá trị biểu thức A tại x=5 là }\dfrac{5}{3}.\\\text{c)Để A=}\dfrac{-1}{3}\\\Rightarrow \dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{-1}{3}\\\Leftrightarrow 3\sqrt{x}=-(\sqrt{x}-2)\\\Leftrightarrow 3\sqrt{x}=2-\sqrt{x}\\\Leftrightarrow 4\sqrt{x}=2\\\Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\\Leftrightarrow x=\dfrac{1}{4}.\text{(thỏa mãn)}\\\text{Vậy với x=}\dfrac{1}{4}\text{ thì A=}\dfrac{-1}{3}.\)