\(A=x^2-6x+11\\ =x^2-2.3.x+3^2+2\\ =\left(x-3\right)^2+2\)
\(\left(x-3\right)^2\ge0\forall x\\ \Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu = xảy ra khi
\(x-3=0\\ \Rightarrow x=3\)
Vậy \(Min_A=2\) khi \(x=3\)
\(A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu"=" xảy ra khi \(x=3\)
\(A=x^2-6x+11\)
\(=x^2-3x-3x+9+2\)
\(=\left(x-3\right)^2+2\)
Mà \(\left(x-3\right)^2+2\ge0\)
Vậy GTLN của A Là 2 khi x=3