$(2x+1)^2-(3x-2)^2+x-11$
$=4x^2+4x+1-(9x^2-12x+4)+x-11$
$=4x^2+5x-10-9x^2+12x-4$
$=(4x^2-9x^2)+(5x+12x)-14$
$=-5x^2+17x-14$
$=-(5x^2-17x+14)$
$=-[(\sqrt{5}x)^2-2.\sqrt{5}x.\dfrac{17\sqrt{5}}{10}+\dfrac{289}{20}-\dfrac{9}{20}]$
$=-(\sqrt{5}x-\dfrac{17\sqrt{5}}{10})^2+\dfrac{9}{20}$
Ta nhận thấy: $-(\sqrt{5}x-\dfrac{17\sqrt{5}}{10})^2\le 0$
$\to -(\sqrt{5}x-\dfrac{17\sqrt{5}}{10})^2+\dfrac{9}{20}\le \dfrac{9}{20}$
$\to$ Dấu "=" xảy ra khi $\sqrt{5}x-\dfrac{17\sqrt{5}}{10}=0$
$\to x=\dfrac{17}{10}$
$\to \max{A}=\dfrac{9}{20}$ khi $x=\dfrac{17}{10}$
