\(\Delta'=m^2-\left(m^2-2m+4\right)=2m-4\ge0\Rightarrow m\ge2\)
Do \(ac=m^2-2m+4=\left(m-1\right)^2+3>0\)
\(\Rightarrow\) Pt có 2 nghiệm cùng dấu
Để pt có 2 nghiệm ko âm thì \(x_1+x_2>0\Rightarrow2m>0\Rightarrow m>0\Rightarrow m\ge2\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-2m+4\end{matrix}\right.\)
\(P=\sqrt{x_1}+\sqrt{x_2}>0\)
\(P^2=x_1+x_2+2\sqrt{x_1x_2}\)
\(P^2=2m+2\sqrt{m^2-2m+4}\)
\(\Rightarrow P=\sqrt{2m+2\sqrt{m^2-2m+4}}\)
Để \(P_{min}\Rightarrow P^2=2\left(m+\sqrt{m^2-2m+4}\right)\) đạt min
Do \(m\ge2\Rightarrow m-2\ge0\Rightarrow m\left(m-2\right)\ge0\)
\(\Rightarrow P^2=2\left(m+\sqrt{m\left(m-2\right)+4}\right)\ge2\left(m+\sqrt{4}\right)=2\left(m+2\right)\ge2.4=8\)
\(\Rightarrow P\ge\sqrt{8}=2\sqrt{2}\)
\(P_{min}=2\sqrt{2}\) khi \(m=2\)
