$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{-x^{2} \ +( \ m-1) x\ +5m\ -\ 6\ =\ 0\ ( 1)}\\ \mathrm{Có\ \vartriangle =( m-1)^{2} +4( 5m-6) =m^{2} +18m-15}\\ \mathrm{Để\ ( 1) \ có\ 2\ nghiệm\ x_{1} \ x_{2} \ \Leftrightarrow \vartriangle \geqslant 0}\\ \mathrm{\Leftrightarrow m^{2} +18m-15\geqslant 0}\\ \mathrm{\Leftrightarrow m\geqslant -9+4\sqrt{6} \ hoặc\ m\leqslant -9-4\sqrt{6}}\\ \mathrm{Theo\ Viet,\ có\ x_{1} +x_{2} =m-1,\ x_{1} x_{2} =6-5m}\\ \mathrm{Ta\ có:\ 4x_{1} +3x_{2} =1,\ thay\ x_{1} =m-1-x_{2} ,\ ta\ được:}\\ \mathrm{4( m-1-x_{2}) +3x_{2} =1}\\ \mathrm{\Leftrightarrow 4m-5=x_{2}}\\ \mathrm{\Rightarrow x_{1} =m-1-4m+5=-3m+4}\\ \mathrm{Ta\ có:\ x_{1} x_{2} =( 4m-3)( -3m+4) =6-5m}\\ \mathrm{\Leftrightarrow -12m^{2} +25m-12=6-5m}\\ \mathrm{\Leftrightarrow -12m^{2} +30m-18=0}\\ \mathrm{\Leftrightarrow m=\frac{3}{2} \ ( TM) \ hoặc\ m=1\ ( TM)}\\ \mathrm{Vậy\ m=\left\{1;\frac{3}{2}\right\}}\\ \\ \end{array}$
