b,
Phương trình có 2 nghiệm khi $\Delta\ge 0$
$\Delta = (2m-1)^2-4(m+3)$
$=4m^2-4m+1-4m-12$
$=4m^2-8m-11\ge 0$
$\Leftrightarrow m\le \dfrac{2-\sqrt{15}}{2}$ hoặc $m\ge \dfrac{2+\sqrt{15}}{2}$
Theo Viet, ta có:
$x_1+x_2=2m-1$
$\Leftrightarrow 2x_1+2x_2=4m-2$
Mà $2x_1+3x_2=13$
$\Rightarrow x_2=13-(4m-2)=15-4m$
$\Rightarrow x_2=2m-1-(15-4m)=6m-16$
Ta có $x_1.x_2=m+3$
$\Rightarrow (15-4m)(6m-16)=m+3$
$\Leftrightarrow -24m^2+154m-240=m+3$
$\Leftrightarrow 24m^2-153m+243=0$
$\Leftrightarrow m=3$ hoặc $m=\dfrac{27}{8}$ (TM)
