Đáp án:
$\rm A_{max} = \dfrac{4}{3} ⇔ x = - \dfrac{1}{3} \\B_{max} = \dfrac{201}{100} ⇔ x = 1$
Giải thích các bước giải:
$\rm a) \\ A = -3x^2 - 2x + 1 \\ = - 3 . ( x^2 + \dfrac{2}{3} x - \dfrac{1}{3} ) \\ = -3 . [ x^2 + 2 . x . \dfrac{1}{3} + (\dfrac{1}{3})^2 - \dfrac{1}{3} - (\dfrac{1}{3})^2) ] \\ = -3 . [ ( x + \dfrac{1}{3} )^2 - \dfrac{4}{9} \\ = -3 . ( x + \dfrac{1}{3} )^2 + \dfrac{4}{3} \leq \dfrac{4}{3} \\ Dấu \ = \ xảy \ ra \ khi \ : \ x + \dfrac{1}{3} = 0 \\ ⇔ x = - \dfrac{1}{3} \\ Vậy \ A_{max} = \dfrac{4}{3} ⇔ x = - \dfrac{1}{3} \\ b) \\ B = \dfrac{2010}{x^2-2x+1001} \\ Ta \ có \ : \ x^2-2x+1001=(x^2-2x+1)+1000 \\ = (x-1)^2+1000 \geq 1000 \\ \to \dfrac{2010}{x^2-2x+1001} \leq \dfrac{2010}{1000}= \dfrac{201}{100} \\ Dấu \ = \ xảy \ ra \ khi \ : \ x - 1 = 0 \\ ⇔ x = 1 \\ Vậy \ B_{max} = \dfrac{201}{100} ⇔ x = 1$
