Để pt có 2 nghiệm ko âm:
\(\left\{{}\begin{matrix}\Delta'=2m-4\ge0\\x_1+x_2=2m\ge0\\x_1x_2=m^2-2m+4\ge0\end{matrix}\right.\) \(\Rightarrow m\ge2\)
Đặt \(Q=\sqrt{x_1}+\sqrt{x_2}\Rightarrow Q^2=x_1+x_2+2\sqrt{x_1x_2}\)
\(Q^2=2m+2\sqrt{m^2-2m+4}\ge2.2+2\sqrt{2^2-2.2+4}=6\)
\(\Rightarrow Q\ge\sqrt{6}\Rightarrow P\ge2018+\sqrt{6}\)
Dấu "=" xảy ra khi \(m=2\)