\(P=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{2}=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\left(2\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right).2}=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)Ta có khi x=64 thì P=\(\dfrac{\sqrt{64}+1}{\sqrt{64}+2}=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
Ta có 2P=\(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}=2-\dfrac{2}{\sqrt{x}+2}\)
Vậy để 2P nhận giá trị nguyên thì \(\sqrt{x}+2\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=-1\\\sqrt{x}+2=1\\\sqrt{x}+2=2\\\sqrt{x}+2=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(l\right)\\\sqrt{x}=-1\left(l\right)\\\sqrt{x}=0\left(n\right)\\\sqrt{x}=-4\left(l\right)\end{matrix}\right.\)\(\Rightarrow x=0\left(tm\right)\)
Vậy khi x=0 thì 2P nhận giá trị nguyên
