\(A=\frac{2x^2-4x+7}{x^2-2x+2}=\frac{2\left(x^2-2x+2\right)+3}{x^2-2x+2}=2+\frac{3}{x^2-2x+2}\)
Ta thấy \(x^2-2x+2=\left(x-1\right)^2+1\ge1\)
\(\Leftrightarrow\frac{3}{x^2-2x+2}\le3\Leftrightarrow A\le5\)
Dấu " = " xảy ra khi x = 1
\(A=\frac{\left(x^2-6x+9\right)-\left(x^2-2x+2\right)}{x^2-2x+2}=\frac{\left(x-3\right)^2}{x^2-2x+2}-1\ge-1\)
Dấu " = " xảy ra khi x = 3
Vậy GTTN của A = - 1 khi x = 3 , GTLN của A = 5 khi x = 1
\(A-5=\frac{2x^2-4x+7}{x^2-2x+2}-5=\frac{2x^2-4x+7-5\left(x^2-2x+2\right)}{\left(x-1\right)^2+1}=\frac{2x^2-4x+7-5x^2+10x-10}{\left(x-1\right)^2+1}\)
\(=\frac{-3x^2+6x-3}{\left(x-1\right)^2+1}=\frac{-3\left(x^2-2x+1\right)}{\left(x-1\right)^2+1}=\frac{-3\left(x-1\right)^2}{\left(x-1\right)^2+1}\le0\)
\(\Rightarrow A-5\le0\Leftrightarrow A\le5\)
GTLN A = 5 khi x = 1
