$7-x^2-3x$
$= -x^2-3x+7$
$= -x^2-3x-\dfrac94+\dfrac{37}4$
$= -\left(x^2+3x+\dfrac94\right)+\dfrac{37}4$
$= -\left(x+\dfrac32\right)^2+\dfrac{37}4$
Vì $\left(x+\dfrac32\right)^2 \ge 0 \; \forall x\in \mathbb{R}$
$⇒ -\left(x+\dfrac32\right)^2 \le 0 \; \forall x\in \mathbb{R}$
$⇒ -\left(x+\dfrac32\right)^2+\dfrac{37}4 \le -\dfrac{19}4\; \forall x\in \mathbb{R}$
Vậy $\max = \dfrac{37}4$ khi $x+\dfrac32 = 0 ⇔ x=-\dfrac32$
