\(P\in Z\Rightarrow3P\in Z\Rightarrow\dfrac{3\sqrt{x}+15}{3\sqrt{x}+1}\in Z\)
\(\Rightarrow1+\dfrac{14}{3\sqrt{x}+1}\in Z\)
\(\Rightarrow3\sqrt{x}+1=Ư\left(14\right)=\left\{1;2;7;14\right\}\) (do \(3\sqrt{x}+1\ge1\))
\(3\sqrt{x}+1=1\Rightarrow x=0\)
\(3\sqrt{x}+1=2\Rightarrow x=\dfrac{1}{9}\notin Z\) (loại)
\(3\sqrt{x}+1=7\Rightarrow x=4\)
\(3\sqrt{x}+1=14\Rightarrow x=\dfrac{169}{9}\notin Z\) (loại)
Thế \(x=\left\{0;4\right\}\) vào P đều thỏa mãn
Vậy ....