Đáp án:
$\left[\begin{array}{l} m=\dfrac{9}{4}\\ m=-\dfrac{2}{3}\end{array} \right.$
Giải thích các bước giải:
$d)Vi-et:x_1+x_2=2m+3 \Rightarrow x_1=2m+3-x_2\\ x_1x_2=4m+2\\ \circledast 2x_1-3x_2=5\\ \Leftrightarrow 2(2m+3-x_2)-3x_2=5\\ \Leftrightarrow 4m+6-2x_2-3x_2=5\\ \Leftrightarrow 4m+1-5x_2=0\\ \Leftrightarrow 4m+1=5x_2\\ \Leftrightarrow x_2=\dfrac{4m+1}{5}\\ \Rightarrow x_1=2m+3-\dfrac{4m+1}{5}=\dfrac{6m+14}{5}\\ x_1x_2=\dfrac{4m+1}{5}.\dfrac{6m+14}{5}=\dfrac{24m^2+62m+14}{25}\\ \text{Mà } x_1x_2=4m+2 \\ \Rightarrow 4m+2 =\dfrac{24m^2+62m+14}{25}\\ \Leftrightarrow 25(4m+2)=24m^2+62m+14\\ \Leftrightarrow 100m+50=24m^2+62m+14\\ \Leftrightarrow 24m^2-38m-36=0\\ \Leftrightarrow \left[\begin{array}{l} m=\dfrac{9}{4}\\ m=-\dfrac{2}{3}\end{array} \right.$
