Đáp án:
Bài 1 :
$(x-3)(x^2+3x+9) =-19$
$⇔x^3 -27 =-19$
$⇔x^3 =-19 +27$
$⇔x^3 = 8$
$⇔x^3 =2^3$
$⇔x=2$
Vậy $x=2$
Bài 2:
$a) A =x^2 +3x-4$
$ = x^2 +2 .x . \dfrac{3}{2} +\dfrac{9}{4} - \dfrac{25}{4}$
$=(x+\dfrac{3}{2})^2 -\dfrac{25}{4}$
Vì $(x+\dfrac{3}{2})^2 ≥ 0$
Nên $(x+\dfrac{3}{2})^2 -\dfrac{25}{4} ≥ -\dfrac{25}{4}$
Dấu ''='' xảy ra khi $x+\dfrac{3}{2} =0⇔x=-\dfrac{3}{2}$
Vậy Min A=$-\dfrac{25}{4}$ tại $x=-\dfrac{3}{2}$
$b) B =-x^2 +5x$
$= -(x^2-5x)$
$=-(x^2 - 2 .x.\dfrac{5}{2} +\dfrac{25}{4} -\dfrac{25}{4})$
$=-(x-\dfrac{5}{2})^2 +\dfrac{25}{4}$
Vì $-(x+\dfrac{5}{2})^2 ≤ 0$
Nên $-(x+\dfrac{5}{2})^2 +\dfrac{25}{4} ≤ \dfrac{25}{4}$
Dấu ''='' xảy ra khi $x-\dfrac{5}{2}=0⇔x=\dfrac{5}{2}$
Vậy Max B $=\dfrac{25}{4}$ tại $x=\dfrac{5}{2}$
