Điều kiện : x \(\ge\) 0
Để P nguyên khi \(\dfrac{10}{\sqrt{x}+2}\) nguyên\(\Rightarrow10⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)Vì \(\sqrt{x}+2\ge2\) với mọi x\(\ge\)0
\(\Rightarrow\sqrt{x}+2\in\left\{2;5;10\right\}\)
-Nếu \(\sqrt{x}+2=5\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
-Nếu \(\sqrt{x}+2=2\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
-Nếu \(\sqrt{x}+2=10\Rightarrow\sqrt{x}=8\Rightarrow x=64\)
Vậy với x\(\in\left\{0;9;64\right\}\)thì P nguyên
Điều kiện : x \(\ge\) 0
Để P nguyên khi \(\dfrac{10}{\sqrt{x}+2}\) nguyên
\(\Rightarrow\)10 \(⋮\) \(\sqrt{x}+2\) \(\Rightarrow\sqrt{x}+2\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Vì \(\sqrt{x}\) + 2 \(\ge\) 2 với mọi x \(_{^{ }\ge}\) 0