\(Q=\dfrac{3x^2-12x+20}{x^2-4x+5}=\dfrac{8\left(x^2-4x+5\right)-5x^2+20x-20}{x^2-4x+5}\)
\(Q=8+\dfrac{-5\left(x^2-4x+4\right)}{x^2-4x+5}\)
\(Q=8+\dfrac{-5\left(x-2\right)^2}{\left(x-2\right)^2+1}\le8\forall x\in R\)
dấu = xảy ra khi \(x-2=0\Leftrightarrow x=2\)
vậy \(Q_{max}=8\) khi x=2