P=\(\dfrac{8\left(2x-1\right)}{x^2+2}=\dfrac{16x-8}{x^2+2}=\dfrac{4x^2+8-4x^2+16x-16}{x^2+2}\)
\(P=\dfrac{4\left(x^2+2\right)-4\left(x^2-4x+4\right)}{x^2+2}=4-\dfrac{4\left(x-2\right)^2}{x^2+2}\le4\)
dấu = xảy ra \(\Leftrightarrow4\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy maxP = 4 khi và chỉ khi x=2