Đáp án:
$a) A =x^2+y^2-3x-3y-2010$
$ = (x^2-2.x.\dfrac{3}{2} + \dfrac{9}{4}) + (y^2 -2. y. \dfrac{3}{2} + \dfrac{9}{4}) -2014,5$
$=(x-\dfrac{3}{2})^2 + (y-\dfrac{3}{2})^2 -2014,5$
Vì $(x-\dfrac{3}{2})^2 + (y-\dfrac{3}{2})^2 ≥ 0$
Nên $(-\dfrac{3}{2})^2 + (y-\dfrac{3}{2})^2 -2014,5 ≥ -2014,5$
Dấu ''='' xảy ra khi $x=y=\dfrac{3}{2}$
Vậy Min A=$-2014,5$ khi $x=y=\dfrac{3}{2}$
$b) B =-x^2-x+2016$
$=-(x^2+x-2016)$
$=-(x^2 +2 . x . \dfrac{1}{2} +\dfrac{1}{4} -2016,25)$
$=-(x+\dfrac{1}{2})^2 +2016,25$
Vì $-(x+\dfrac{1}{2})^2 ≤ 0$
Nên $-(x+\dfrac{1}{2})^2 +2016,25 ≤ 2016,25$
Dấu ''='' xảy ra khi $x+\dfrac{1}{2} =0⇔x=-\dfrac{1}{2}$
Vậy Max $B =2016,25$ tại $x=-\dfrac{1}{2}$
