\(A=\frac{2+2x-x^2}{x^2-2x+3}=\frac{-x^2+2x-3+5}{x^2-2x+3}=-1+\frac{5}{x^2-2x+3}\)
\(x^2-2x+3=\left(x-1\right)^2+2\ge2\forall x\)
=>\(\frac{5}{x^2-2x+3}\le=\frac52\forall x\)
=>\(A=\frac{5}{x^2-2x+3}-1\le\frac52-1=\frac32\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1