\(P+1=\dfrac{8x+1}{4x^2+3}+1=\dfrac{8x+1+4x^2+3}{4x^2+3}=\dfrac{4\left(x+1\right)^2}{4x^2+3}\ge0\)\(P+1\ge0\Rightarrow P\ge-1\) tại x =-1
\(P-\dfrac{4}{3}=\dfrac{8x+1}{4x^2+3}-\dfrac{4}{3}=\dfrac{3.\left(8x+1\right)-4\left(4x^2+3\right)}{4x^2+3}=\dfrac{-\left(4x-3\right)^2}{4x^2+3}\le0\)
\(P-\dfrac{4}{3}\le0\Rightarrow P\le\dfrac{4}{3}\) khi x =3/4