$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{a.\ Khi\ x_{1} =2,\ PT\ trở\ thành:}\\ \mathrm{2^{2} -2( 2m+1) .2+4m^{2} +m-3=0}\\ \mathrm{\Leftrightarrow 4m^{2} -7m-3=0}\\ \mathrm{\Leftrightarrow m=\dfrac{7\pm \sqrt{97}}{8}}\\ \mathrm{Theo\ Viet,\ có\ x_{1} +x_{2} =2( 2m+1)}\\ \mathrm{\Leftrightarrow x_{2} =4m+2-x_{1} =4m}\\ \mathrm{\Leftrightarrow x_{2} =\dfrac{7\pm \sqrt{97}}{2}}\\ \mathrm{b.\ Để\ PT\ có\ 2\ nghiệm\ phân\ biệt}\\ \mathrm{\Leftrightarrow \Delta '=( 2m+1)^{2} -4m^{2} +m-3 >0}\\ \mathrm{\Leftrightarrow 5m-2 >0}\\ \mathrm{\Leftrightarrow m >\dfrac{2}{5}}\\ \mathrm{Vậy\ m>\dfrac25} \end{array}$
