a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(F=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c) Để F dương thì F>0
\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}>0\)
mà \(\sqrt{x}\ge0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4(nhận)
Vậy: để F dương thì \(x>4\)
d) Để F=3 thì \(\frac{\sqrt{x}}{\sqrt{x}-2}=3\)
\(\Leftrightarrow\sqrt{x}=3\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow\sqrt{x}=3\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}-3\sqrt{x}=-6\)
\(\Leftrightarrow-2\sqrt{x}=-6\)
\(\Leftrightarrow\sqrt{x}=3\)
hay x=9(nhận)
Vậy: để F=3 thì x=9
e) Để \(F\in Z\) thì \(\sqrt{x}⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2+2⋮\sqrt{x}-2\)
mà \(\sqrt{x}-2⋮\sqrt{x}-2\forall x\) thỏa mãn ĐKXĐ
nên \(2⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;-2;-1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;4;0;1\right\}\)
hay \(x\in\left\{9;16;0;1\right\}\)(nhận)
Vậy: để \(F\in Z\) thì \(x\in\left\{9;16;0;1\right\}\)