$N=2x-2x^2-5$
$↔2.(x-x^2-\dfrac{5}{2})$
$↔-2.(x^2-x+\dfrac{5}{2})$
$↔-2.(x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{9}{4})$
$↔-2.(x-\dfrac{1}{2})^2-\dfrac{9}{2}$
Ta thấy: $(x-\dfrac{1}{2})^2≥0$
$→-2(x-\dfrac{1}{2})^2≤0$
$→-2(x-\dfrac{1}{2})-\dfrac{9}{2}≤-\dfrac{9}{2}$
$→$ Dấu "=" xảy ra khi $x-\dfrac{1}{2}=0$
$→x=\dfrac{1}{2}$
$→N_{max}=-\dfrac{9}{2}$
Vậy $N_{max}=-\dfrac{9}{2}$ khi $x=\dfrac{1}{2}$
