\(A=x^2-20x+5\)
\(=\left(x^2-2.x.10+10^2\right)-20+5\)
\(=\left(x-10\right)^2-15\)
Do \(\left(x-10^2\right)\ge0\) \(\Rightarrow\left(x-10\right)^2-15\ge-15\)
Dấu = xảy ra \(\Leftrightarrow\left(x-10\right)^2=0\) \(\Leftrightarrow x-10=0\Leftrightarrow x=10\)
Vậy \(A_{min}=-15\Leftrightarrow x=10\)