\(\Delta=\left(m+3\right)^2-8m=m^2-2m+9=\left(m-1\right)^2+8>0\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm pb
Theo Viet ta có \(\left\{{}\begin{matrix}x_1+x_2=\frac{m+3}{2}\\x_1x_2=\frac{m}{2}\end{matrix}\right.\)
\(A=\left|x_1-x_2\right|\ge0\)
\(\Leftrightarrow A^2=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(\Leftrightarrow A^2=\left(\frac{m+3}{2}\right)^2-\frac{4m}{2}\)
\(\Leftrightarrow4A^2=m^2-2m+9\)
\(\Leftrightarrow4A^2=\left(m-1\right)^2+8\ge8\)
\(\Rightarrow A^2\ge2\Rightarrow A\ge\sqrt{2}\)
\(\Rightarrow A_{min}=\sqrt{2}\) khi \(m=1\)