Đáp án:
$3)\\ a)min_A=\dfrac{19}{4} \Leftrightarrow x=-\dfrac{5}{2}\\ b)max_B=-3\Leftrightarrow x=1.$
Giải thích các bước giải:
$3)\\ a)A=x^2+5x+11\\ =x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+\dfrac{19}{4}\\ =\left(x+\dfrac{5}{2}\right)^2+\dfrac{19}{4} \ge \dfrac{19}{4} \ \forall \ x$
Dấu "=" xảy ra $ \Leftrightarrow x+\dfrac{5}{2}=0 \Leftrightarrow x=-\dfrac{5}{2}$
Vậy $min_A=\dfrac{19}{4} \Leftrightarrow x=-\dfrac{5}{2}$
$b)B=4x-2x^2-5\\ =-2\left(x^2-2x+\dfrac{5}{2}\right)\\ =-2\left(x^2-2x+1+\dfrac{3}{2}\right)\\ =-2\left(x^2-2x+1\right)-3\\ =-2(x-1)^2-3 \le -3 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-1=0 \Leftrightarrow x=1$
Vạy $max_B=-3\Leftrightarrow x=1.$
