\(\Delta=m^2-12>0\Rightarrow\left[{}\begin{matrix}m>2\sqrt{3}\\m< -2\sqrt{3}\end{matrix}\right.\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=3\end{matrix}\right.\)
Do \(x_1x_2=3>0\Rightarrow x_1;x_2\) luôn cùng dấu.
Th1: \(m>2\sqrt{3}\Rightarrow x_1+x_2>0\Rightarrow x_1;x_2>0\)
\(\dfrac{1}{x_1}-\dfrac{1}{x_2}=\dfrac{2}{3}\Rightarrow x_2-x_1=\dfrac{2}{3}x_1x_2=2\)
\(\Rightarrow x_2=x_1+2\)
Thế vào \(x_1x_2=3\Rightarrow x_1\left(x_1+2\right)=3\Rightarrow x_1^2+2x_1-3=0\Rightarrow\left[{}\begin{matrix}x_1=1\\x_1=-3\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x_2=3\Rightarrow m=x_1+x_2=4\) (thỏa mãn)
TH2: \(m< -2\sqrt{3}\Rightarrow x_1;x_2< 0\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{2}{3}\Rightarrow x_1+x_2=\dfrac{2}{3}x_1x_2=2\)
\(\Rightarrow m=2>-2\sqrt{3}\) (ktm)
