$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ m=1,\ ( 1) \ trở\ thành:\\ x^{2} -2x-3=0\\ \Leftrightarrow x=3\ or\ x=-1\\ Vậy\ m=1;\ S=\{-1;3\}\\ 2.\ \Delta '=( m-2)^{2} -m^{2} +4m=4 >0\\ \Rightarrow ( 1) \ luôn\ có\ 2\ nghiệm\ phân\ biêt:\\ x_{1} =-( m-2) +2=4-m\\ x_{2} =-m\\ Ta\ có:\ \frac{3}{x_{1}} +x_{2} =\frac{3}{x_{2}} +x_{1} \ ( m\neq 0;\ m\neq 4)\\ \Leftrightarrow \frac{3}{4-m} +m =\frac{3}{m} +4-m\\ \Leftrightarrow \frac{3m-3.4+3m}{4m-m^{2}} =4-2m\\ \Leftrightarrow 6( m-2) =2m( 4-m)( 2-m)\\ \Leftrightarrow m=2\ ( TM) ;\ m=2\pm \sqrt{7} \ ( TM)\\ \\ \end{array}$
