Ta có: \(A=\frac{2m^2-4m+5}{m^2-2m+2}\)
\(=\frac{2m^2-4m+2+3}{m^2-2m+1+1}=\frac{2\left(m^2-2m+1\right)+3}{\left(m^2-2m+1\right)+1}\)
\(=\frac{2\left(m-1\right)^2+3}{\left(m-1\right)^2+1}\ge\frac{3}{1}=3\) (do \(\left(m-1\right)^2\ge0\))
Dấu "=" xảy ra \(\Leftrightarrow m-1=0\Leftrightarrow m=1\)
Vậy \(A_{min}=3\Leftrightarrow m=1\)
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\(A=2+\frac{1}{m^2-2m+1+1}=2+\frac{1}{\left(m-1\right)^2+1}\)
\(\left(m-1\right)^2+1\ge1\Leftrightarrow\frac{1}{\left(m-1\right)^2+1}\le1\)
\(\Rightarrow A\le3\)
\("="\Leftrightarrow m=1\)