\(x^2-\left(2m+1\right)x+m^2+m+1=0\)
2 nghiệm phân biệt khi
\(\Delta=\left(2m+1\right)^2-4\left(m^2+m+1\right)=0\)
=>\(\Delta=4m^2+4m+1-4m^2-4m-4=0\)
=>\(\Delta=-3< 0\)
b)\(\orbr{\begin{cases}x_1=\frac{2m+1-3}{2}=\frac{2m+1}{2}-\frac{3}{2}\\x_2=\frac{2m+1+3}{2}=\frac{2m+1}{2}+\frac{3}{2}\end{cases}}\)
\(x_1-x_2=-3\)