\(x^2-4x+m-3=0\left(1\right)\)
\(\Delta'=b'^2-ac=\left(-2\right)^2-\left(m-3\right)=4-m+3=7-m\)
Để pt (1) có 2 nghiệm phân biệt \(x_1;x_2\) thì \(\Delta'>0\Leftrightarrow7-m>0\Leftrightarrow m< 7\)
Theo Viet ta có:
\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1\cdot x_2=m-3\end{matrix}\right.\)
Ta có:
\(\left|x_1-x_2\right|=3\Leftrightarrow\sqrt{\left(x_1-x_2\right)^2}=3\\ \Leftrightarrow\left(x_1-x_2\right)^2=9\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=9\\ \Leftrightarrow4^2-4\left(m-3\right)-9=0\\ \Leftrightarrow-4m+19=0\\ \Leftrightarrow m=\frac{19}{4}\left(tm\right)\)
Vậy .................