\(A=m^2-2m-5\)
\(=m^2-2m+1-6\)
\(=\left(m-1\right)^2-6\ge-6\)
Dấu '' = '' xảy ra khi \(\left(m-1\right)^2=0\Leftrightarrow m=1\)
Vậy \(Min_A=-6\) khi \(m=1\)
\(A=m^2-2m-5\)
\(=\left(m^2-2m+1\right)-6\)
\(=\left(m-1\right)^2-6\ge-6\left(Vì\left(m-1\right)^2\ge0\forall m\right)\)
Min \(A=-6\Leftrightarrow m=1\)