Đáp án:

a) $A=\dfrac{\sqrt{x}}{\sqrt{x}-2}$

b) $x=\dfrac{1}{4}$ thì $A=-\dfrac{1}{3}$

c) $x=9$ hoặc $x=1$ hoặc $x=16$ hoặc $x=0$ thì A nhận giá trị nguyên

Giải thích các bước giải:

a)

$A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\,\,\,(x\ge0;x\ne 4)\\=\dfrac{x}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\=\dfrac{x}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\=\dfrac{x+2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}}{\sqrt{x}-2}$

b)

$A=\dfrac{\sqrt{x}}{\sqrt{x}-2}\,\,\,(x\ge0;x\ne4)$

Để $A=-\dfrac{1}{3}\to \dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\\\to 3\sqrt{x}=2-\sqrt{x}\\\to 4\sqrt{x}=2\\\to \sqrt{x}=\dfrac{1}{2}\\\to x=\dfrac{1}{4}$ (thoả mãn)

Vậy $x=\dfrac{1}{4}$ thì $A=-\dfrac{1}{3}$

c)

$A=\dfrac{\sqrt{x}}{\sqrt{x}-2}\,\,\,(x\ge0;x\ne4)$

$A=\dfrac{(\sqrt{x}-2)+2}{\sqrt{x}-2}\\=1+\dfrac{2}{\sqrt{x}-2}$

Để A có giá trị nguyên

$\to 2\vdots(\sqrt{x}-2)\to \sqrt{x}-2\in Ư(2)=\{\pm1;\pm2\}\\\to \left[\begin{array}{l}\sqrt{x}-2=1\\\sqrt{x}-2=-1\\\sqrt{x}-2=2\\\sqrt{x}-2=-2\end{array}\right.\\\to \left[\begin{array}{l}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=4\\\sqrt{x}=0\end{array}\right.\\\to \left[\begin{array}{l}x=9\\x=1\\x=16\\x=0\end{array}\right.$ (thoả mãn)

Vậy $x=9$ hoặc $x=1$ hoặc $x=16$ hoặc $x=0$ thì A nhận giá trị nguyên