PT: \(x^2-2mx+m^2-m+1=0\)
Ta có: \(\Delta'=m-1\)
Để phương trình có 2 nghiệm \(\Leftrightarrow\Delta'\ge0\) \(\Leftrightarrow m\ge1\)
Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1\cdot x_2=m^2-m+1\end{matrix}\right.\)
\(P=x_1^2+x^2_2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(\Rightarrow P=2m^2+2m-2\) \(=2\left(m^2+m-1\right)=2\left(m^2-2m+1+3m-2\right)\) \(=2\left(m-1\right)^2+6m-4\ge6m-4\ge2\)
Dấu bằng xảy ra \(\Leftrightarrow m-1=0\Leftrightarrow m=1\) (TM)
Vậy \(P_{Min}=2\) khi \(m=1\)