\(A=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)
Dấu "=" xảy ra khi: \(x=2\)
Ta có:\(A=x^2-4x+x5\)
\(\Rightarrow A=x^2-2.x.2+2^2-4+5\)
\(\Rightarrow A=\left(x-2\right)^2+1\)
Do \(\left(x-2\right)^2\ge0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=2\))
\(\Rightarrow\left(x-2\right)^2+1\ge1\) hay \(A\ge1\) (dấu "=" xảy ra \(\Leftrightarrow x=2\))
Vậy \(A_{min}=1\) tại \(x=2\)