1: \(\Delta=\left\lbrack-2\left(m-1\right)\right\rbrack^2-4\cdot1\cdot\left(m-2\right)\)
\(=4m^2-8m+4-4m+8=4m^2-12m+12\)
\(=4m^2-12m+9+3=\left(2m-3\right)^2+3>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m-1\right);x_1x_2=\frac{c}{a}=m-2\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(2m-2\right)^2-2\left(m-2\right)=4m^2-8m+4-2m+4=4m^2-10m+8\)
\(x_1+x_2+\sqrt{x_1^2+x_2^2}=2\)
=>\(2m-2+\sqrt{4m^2-10m+8}=2\)
=>\(\sqrt{4m^2-10m+8}=2+2-2m=-2m+4\)
=>\(\begin{cases}-2m+4\ge0\\ \left(-2m+4\right)^2=4m^2-10m+8\end{cases}\Rightarrow\begin{cases}-2m\ge-4\\ 4m^2-16m+16-4m^2+10m-8=0\end{cases}\)
=>\(\begin{cases}m\le2\\ -6m+8=0\end{cases}\Rightarrow m=\frac43\)
