\(Q=\frac{1}{x^2-2x+3}=\frac{1}{\left(x^2-2x+1\right)+2}=\frac{1}{\left(x-1\right)^2+2}\)
Để \(\frac{1}{\left(x-1\right)^2+2}\) max <=> \(\left(x-1\right)^2+2\) min
Mà \(\left(x-1\right)^2+2\ge2\) \(\forall x\)
\(\Rightarrow Q=\frac{1}{\left(x-1\right)^2+2}\ge\frac{1}{2}\)
Dấu "=" xảy ra <=> \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Q_{MAX}=\frac{1}{2}\) tại \(x=1\)