Đặt \(y=\frac{3x^2+2x+1}{x^2-2x+3}\Rightarrow y.x^2-2yx+3y=3x^2+2x+1\)
\(\Leftrightarrow\left(y-3\right)x^2-2\left(y+1\right)x+3y-1=0\)
\(\Delta'=\left(y+1\right)^2-\left(y-3\right)\left(3y+1\right)\ge0\)
\(\Leftrightarrow-y^2+5y+2\ge0\)
\(\Rightarrow\frac{5-\sqrt{33}}{2}\le y\le\frac{5+\sqrt{33}}{2}\)