$x^2-(3m+1)x+2m^2+m-1=0\\\text{Ta có: }\Delta=(3m+1)^2-4(2m^2+m-1)\\\hspace{2cm}=9m^2+6m+1-8m^2-4m+4\\\hspace{2cm}=m^2+2m+5\\\hspace{2cm}=(m+1)^2+4>0\\\to \text{Phương trình có 2 nghiệm phân biệt }\\\text{Theo Viet: }\begin{cases}x_1+x_2=-\dfrac{b}{a}=3m+1\\x_1x_2=\dfrac{c}{a}=2m^2+m-1\end{cases}\\\text{Theo đề bài: }x_1^2+x_2^2-3x_1x_2=-\dfrac{3}{2}\\⇔(x_1+x_2)^2-5x_1x_2=-\dfrac{3}{2}\\⇔(3m+1)^2-5(2m^2+m-1)=-\dfrac{3}{2}\\⇔9m^2+6m+1-10m^2-5m+5=-\dfrac{3}{2}\\⇔-m^2+m+\dfrac{15}{2}=0\\⇔m=\dfrac{1\pm\sqrt{31}}{2}$
