$=\frac{2016}{(2x-\frac{1}{4})^2+\frac{21}{16}}$
Có: $(2x-\frac{1}{4})^2 \geq 0$ => $(2x-\frac{1}{4})^2+\frac{21}{16} \geq \frac{21}{16}$
=> $=\frac{2016}{(2x-\frac{1}{4})^2+\frac{21}{16}}\leq \frac{2016}{\frac{21}{16}}=1536$
Vậy Max H=1536 , dấu "='' tại $x=\frac{1}{8}$