$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{\ x^{2} \ -\ 2\ ( m\ +1) x\ +m-4=0\ ( 1)}\\ \mathrm{a.\ m=1,\ PT\ trở\ thành:}\\ \mathrm{x^{2} -4x-3=0}\\ \mathrm{\Leftrightarrow x=2\pm \sqrt{7}}\\ \mathrm{b.\ \vartriangle '=( m+1)^{2} -m+4=m^{2} -m+5=\left( m-\frac{1}{2}\right)^{2} +\frac{19}{4} >0}\\ \mathrm{\Rightarrow ( 1) \ luôn\ 2\ nghiệm\ phân\ biệt\ x_{1} ,x_{2}}\\ \mathrm{Theo\ Viet\ có:\ x_{1} +x_{2} =2m+2\ ( *) \ \ \ \ \ \ \ \ }\\ \mathrm{x_{1} x_{2} =m-4( **) \Rightarrow m=x_{1} x_{2} +4,\ thay\ vào\ ( *) :}\\ \mathrm{x_{1} +x_{2} =2( x_{1} x_{2} +4) +2=2x_{1} x_{2} +10}\\ \mathrm{Vậy\ x_{1} +x_{2} =2x_{1} x_{2} +10} \end{array}$
