\(P=\left(\dfrac{\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+2\right)}+\dfrac{\left(6-3\sqrt{x}\right)}{\left(\sqrt{x}-2\right)}\right)\left(x-4\right)\left(dk:x\ne4,x\ge0\right)\\ =\left(\dfrac{\left(\sqrt{x}+6\right)\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x-4}\right)\left(x-4\right)\)
\(=x-2\sqrt{x}+6\sqrt{x}-12-3\left(x-4\right)\\ =x+4\sqrt{x}-12-3x+12\)
\(=-2x+4\sqrt{x}\)
Để P là một số nguyên dương thì \(P>0\Leftrightarrow-2x+4\sqrt{x}>0\\ \Leftrightarrow-2\sqrt{x}\left(\sqrt{x}-2\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}-2\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 0\left(ktm\right)\\x>4\left(tm\right)\end{matrix}\right.\)
Vậy \(x>4\left(dk:x\in Z\right)\) thì P là 1 số nguyên dương.
