\(P=A.B\in Z\\ \Leftrightarrow P=\dfrac{\sqrt{x}+10}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}+2}\in Z\\ \Leftrightarrow P=\dfrac{\sqrt{x}+10}{\sqrt{x}+2}\in Z\\ \Leftrightarrow\sqrt{x}+10⋮\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}+2+8⋮\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\Rightarrow x=\left\{0;4;36\right\}\)
\(P=A.B=\dfrac{\sqrt{x}+10}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{\sqrt{x}+10}{\sqrt{x}+2}=1+\dfrac{8}{\sqrt{x}+2}\left(đk:x>0\right)\)
Để P nguyên thì\(8⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
Vì \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}\in\left\{2;6\right\}\)
\(\Rightarrow x\in\left\{4;36\right\}\)
Ta có: P=AB
\(=\dfrac{\sqrt{x}+10}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+10}{\sqrt{x}+2}\)
Để P nguyên thì \(\sqrt{x}+10⋮\sqrt{x}+2\)
\(\Leftrightarrow8⋮\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}+2\in\left\{2;4;8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;6\right\}\)
hay \(x\in\left\{4;36\right\}\)
