a. Với m=6, ta có pt: \(x^2-5x+6=0\)
\(\Delta=25-4.6=1>0\)
Vậy pt có 2 nghiệm phân biệt.
\(x_1=\dfrac{5+1}{2}=3\)
\(x_2=\dfrac{5-1}{2}=2\)
b. Có : \(x_1+x_2=\dfrac{-\left(-5\right)}{1}=5,x_1x_2=\dfrac{m}{1}=m\)
\(\left|x_1-x_2\right|=3\)
\(\Leftrightarrow\sqrt{x_1^2+x_2^2-2x_1x_2}=3\)
\(\Leftrightarrow\sqrt{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-2m}=3\)
\(\Leftrightarrow\sqrt{25-2m-2m}=3\)(ĐK:\(x\le\dfrac{25}{4}\)
\(\Leftrightarrow25-4m=9\)
\(\Leftrightarrow m=4\)(TM).
Vậy m=4.