Xét hiệu : \(Q-2=\frac{3x^2-4x+6}{x^2+1}-2\)
\(=\frac{3x^2-4x+6}{x^2+1}-\frac{2\left(x^2+1\right)}{x^2+1}\)
\(=\frac{x^2-4x+4}{x^2+1}=\frac{\left(x-2\right)^2}{x^2+1}\)
Ta thấy : \(\left(x-2\right)^2\ge0\forall x,x^2+1>0\)
\(\Rightarrow\frac{\left(x-2\right)^2}{x^2+1}\ge0\forall x\)
hay : \(Q-2\ge0\forall x\Rightarrow Q\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
Vậy : min \(Q=2\) tại \(x=2\)