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