\(\Delta=\left(m+1\right)^2-4\left(m^2-2m+2\right)\ge0\)
\(\Leftrightarrow-3m^2+10m-7\ge0\Leftrightarrow1\le m\le\frac{7}{3}\)
Khi đó pt có 2 nghiệm thỏa: \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m^2-2m+2\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(A=\left(m+1\right)^2-2\left(m^2-2m+2\right)\)
\(A=-m^2+6m-3\)
\(A=-m^2+6m-5+2=\left(m-1\right)\left(5-m\right)+2\ge2\) ; \(\forall m\in\left[1;\frac{7}{3}\right]\)
\(A_{min}=2\) khi \(m=1\)
\(A=-m^2+6m-\frac{77}{9}+\frac{50}{9}=\left(m-\frac{7}{3}\right)\left(\frac{11}{3}-m\right)+\frac{50}{9}\le\frac{50}{9}\)
\(A_{max}=\frac{50}{9}\) khi \(m=\frac{7}{3}\)
