\(\Delta=\left(2m+1\right)^2-4m-4=4m^2-3\ge0\)
Theo Viet \(\left\{{}\begin{matrix}x_1+x_2=2m+1\\x_1x_2=m+1\end{matrix}\right.\)
Để đề bài xác định \(\Rightarrow m\ne-\frac{1}{2}\)
\(A=\frac{\left(x_1+x_2\right)^2-4x_1x_2}{\left(x_1+x_2\right)^2}=\frac{\left(2m+1\right)^2-4\left(m+1\right)}{\left(2m+1\right)^2}=\frac{4m^2-3}{\left(2m+1\right)^2}\)
\(A=\frac{8m^2-6}{2\left(2m+1\right)^2}=\frac{12m^2+12m+3-4m^2-12m-9}{2\left(2m+1\right)^2}=\frac{3\left(2m+1\right)^2-\left(2m+3\right)^2}{2\left(2m+1\right)^2}\)
\(A=\frac{3}{2}-\frac{\left(2m+3\right)^2}{\left(2m+1\right)^2}\le\frac{3}{2}\)
\(\Rightarrow A_{max}=\frac{3}{2}\) khi \(m=-\frac{3}{2}\) (thỏa mãn)
