Phương trình có $2$ nghiệm phân biệt khi $Δ>0$
$↔ 9-4.2.2m>0$
$↔ 16m<9$
$↔ m<\dfrac{9}{16}$
Theo định lí $Vi-ét$:
$x_{1}+x_{2}=\dfrac{3}{2}$
$x_{1}x_{2}=m$
$→ \Bigg|\dfrac{1}{x_{1}}-\dfrac{1}{x_{2}}\Bigg|=1$
$↔ \Bigg|\dfrac{x_{2}-x_{1}}{x_{1}x_{2}}\Bigg|=1$
$↔ \sqrt[]{(x_{2}-x_{1})^2}=m$
$↔ \sqrt[]{(x_{1}+x_{2})^2-4x_{1}x_{2}}=m$
$↔ \sqrt[]{\dfrac{9}{4}-4m}=m$
$→ m^2+4m-\dfrac{9}{4}=0$
$↔ \left[ \begin{array}{l}m=\dfrac{1}{2}\\x=-\dfrac{9}{2}\end{array} \right.$ (thỏa mãn)