\(H=\dfrac{x^2-6x+1}{x^2+1}=\dfrac{4x^2+4-3x^2-6x-3}{x^2+1}\)
\(=\dfrac{4\left(x^2+1\right)-3\left(x^2+2x+1\right)}{x^2+1}=4-\dfrac{3\left(x+1\right)^2}{x^2+1}\)
Ta có: \(\dfrac{3\left(x+1\right)^2}{x^2+1}\ge0\forall x\Rightarrow H=4-\dfrac{3\left(x+1\right)^2}{x^2+1}\le4\forall x\)
\(\Rightarrow H_{max}=4\Leftrightarrow x+1=0\Leftrightarrow x=-1\)