\(\Delta=25-4m\ge0\Rightarrow m\le\dfrac{25}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m\end{matrix}\right.\)
\(\left|x_1-x_2\right|=3\Leftrightarrow\left(x_1-x_2\right)^2=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=9\)
\(\Leftrightarrow25-4m=9\Rightarrow m=4\) (thỏa mãn)
Pt có 2 nghiệm
\(\to \Delta=(-5)^2-4.1.m=25-4m\ge 0\\\leftrightarrow 4m\le 25\\\leftrightarrow m\le\dfrac{25}{4}\)
Theo Viét
\(\begin{cases}x_1+x_2=5\\x_1x_2=m\end{cases}\)
\(|x_1-x_2|=3\\\leftrightarrow \sqrt{(x_1-x_2)^2}=3\\\leftrightarrow \sqrt{x_1^2+x_2^2-2x_1x_2}=3\\\leftrightarrow \sqrt{(x_1+x_2)^2-4x_1x_2}=3\\\leftrightarrow \sqrt{5^2-4m}=3\\\leftrightarrow 25-4m=9\\\leftrightarrow 4m=16\\\leftrightarrow m=4(TM)\)
Vậy \(m=4\) thỏa mãn hệ thức