\(\Delta=m^2-4m+4=\left(m-2\right)^2\ge0\)
\(\Rightarrow\) pt đã cho luôn có 2 nghiệm
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)
\(B=\frac{2x_1x_2+3}{x_1^2+x_2^2+2+2x_1x_2}=1\)
\(\Leftrightarrow\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=1\)
\(\Leftrightarrow\frac{2m+1}{m^2+2}=1\)
\(\Leftrightarrow m^2-2m+1=0\Rightarrow m=1\)