Đáp án:
$A = 9x^2+8x-5$
$ = (3x)^2 +2.3x.\dfrac{4}{3} + \dfrac{16}{9} -\dfrac{61}{9}$
$ = (3x +\dfrac{4}{3})^2 - \dfrac{61}{9}$
Vì $(3x+\dfrac{4}{3})^2 ≥ 0$
Nên $(3x+\dfrac{4}{3})^ -\dfrac{61}{9} ≥ -\dfrac{61}{9}$
Dấu ''='' xảy ra khi $3x+\dfrac{4}{3} =0⇔x=-\dfrac{4}{9}$
Vậy Min A $=-\dfrac{61}{9}$ tại $x=-\dfrac{4}{9}$
$b =-3x^2 -2x+1$
$ = -(3x^2+2x-1)$
$ = -[(\sqrt[]{3}x)^2 + 2 . \sqrt[]{3}x . \dfrac{\sqrt[]{3}}{3} +\dfrac{1}{3} -\dfrac{4}{3} ]$
$ =-(\sqrt[]{3}x +\dfrac{\sqrt[]{3}}{3})^2 +\dfrac{4}{3}$
Vì $-(\sqrt[]{3}x+\dfrac{\sqrt[]{3}}{3})^2≤ 0$
NÊN $-(\sqrt[]{3}x +\dfrac{\sqrt[]{3}}{3})^2 +\dfrac{4}{3} ≤ \dfrac{4}{3}$
Dấu ''='' xảy ra khi $\sqrt[]{3}x +\dfrac{\sqrt[]{3}}{3} =0⇔x=-\dfrac{1}{3}$
Vậy Max B $=\dfrac{4}{3}$ tại $x=-\dfrac{1}{3}$