\(A\left(x\right)=\dfrac{2x^2-6x+5}{x^2-2x+1}=\dfrac{2x^2-6x+5}{\left(x-1\right)^2}\)
Đặt \(x-1=t\ne0\Rightarrow x=t+1\)
\(\Rightarrow A=\dfrac{2\left(t+1\right)^2-6\left(t+1\right)+5}{t^2}=\dfrac{2t^2-2t+1}{t^2}\)
\(A=2-\dfrac{2}{t}+\dfrac{1}{t^2}=\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
\(A_{min}=1\) khi \(\dfrac{1}{t}-1=0\Rightarrow t=1\Rightarrow x=2\)