\(x^2\ge0\)
\(\Rightarrow2015x\ge0\)
\(\Rightarrow1-x^2\ge1\)
\(\Rightarrow\sqrt{1-x^2}\ge1\)
\(\Rightarrow\dfrac{2017-2015x}{\sqrt{1-x^2}}\ge\dfrac{2017}{1}=2017\)
Dấu "=" xảy ra khi \(x^2=0\)
\(\Leftrightarrow x=0\)
Vậy \(P\min\limits=2017\Leftrightarrow x=0\)