Ta có: \(A=m\left(x+m+1\right)=m^2+2x-2\)
\(\Leftrightarrow\left(m-2\right)x=\left(m-2\right)\left(m^2+m+1\right)\)
Vì: \(m\ne2\) nên:
\(\Rightarrow x=m^2+m+1=\left(m+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow m=-\frac{1}{2}\left(tmđk\right)\)
\(\Rightarrow Min_A=\frac{3}{4}\)