Đáp án:
$A = 2x^2-8x-11$
$ = (\sqrt[]{2}x)^2 -2 . \sqrt[]{2} . 2\sqrt[]{2} + 8 -19$
$ = (\sqrt[]{2}x - 2\sqrt[]{2})^2 -19$
Vì $(\sqrt[]{2}x - 2\sqrt[]{2})^2 ≥ 0$
Nên $(\sqrt[]{2}x-2\sqrt[]{2})^2 - 19 ≥ -19$
Dấu ''='' xảy ra khi $\sqrt[]{2}x - 2\sqrt[]{2} =0⇔ x = 2$
Vậy Min A $=-19$ tại $x=2$
$B = 9-15x-3x^2$
$ = -(3x^2+15x-9)$
$ = -[(\sqrt[]{3}x)^2 + 2 . \sqrt[]{3}x . \dfrac{5\sqrt[]{3}}{2} + \dfrac{75}{4} - \dfrac{111}{4} ]$
$ = -(\sqrt[]{3}x +\dfrac{5\sqrt[]{3}}{2})^2 +\dfrac{111}{4}$
Vì $-(\sqrt[]{3}x +\dfrac{5\sqrt[]{3}}{2})^2 ≤ 0$
Nên $-(\sqrt[]{3}x +\dfrac{5\sqrt[]{3}}{2})^2 + \dfrac{111}{4} ≤ \dfrac{111}{4}$
Dấu ''='' xảy ra khi $\sqrt[]{3}x + \dfrac{5\sqrt[]{3}}{2} =0⇔ x = -\dfrac{5}{2}$
Vậy Max B $=\dfrac{111}{4}$ tại $x=-\dfrac{5}{2}$
