Đặt
\(A=x^2-10x+1\\ =x^2-10+25-24\\ =\left(x-5\right)^2+\left(-24\right)\\ \left(x-5\right)^2\ge0\forall x\\ \Rightarrow\left(x-5\right)^2+\left(-24\right)\ge-24\forall x\)
Dấu "=" xảy ra khi \(\left(x-5\right)^2=0\Rightarrow x-5=0\Rightarrow x=5\)
Vậy \(Min_A=-24\) khi \(x=5\)