a) $B = b- \dfrac{9b^2}{25}$
$\to - B = \dfrac{9b^2}{25} - b$
$ = \dfrac{9b^2}{25} - 2.\dfrac{3b}{5}.\dfrac{5}{6} + \dfrac{25}{36} - \dfrac{25}{36}$
$ = \bigg(\dfrac{3b}{5} - \dfrac{5}{6}\bigg)^2 - \dfrac{25}{36} ≥ - \dfrac{25}{36}$
$\to B ≤ \dfrac{25}{36}$
Dấu "=" xảy ra $⇔b=\dfrac{25}{18}$
b) $C = 2x-x^2-4$
$ = -(x^2-2x+1) - 3$
$ = -(x-1)^2 - 3 ≤ - 3$
Dấu "=" xảy ra $⇔x=1$
c) $D = -x^2-4x$
$ = -(x^2+4x+4) + 4$
$ = -(x+2)^2 + 4 ≤ 4$
Dấu "=" xảy ra $⇔x=-2$
d) $E = -9x^2+24x-18$
$ = -(9x^2-24x) - 18$
$ = -[(3x)^2-2.3x.4+16] - 18 + 16$
$ = -(3x-4)^2 -2 ≤ -2$
Dấu "=" xảy ra $⇔x=\dfrac{4}{3}$
e) $D = 4x-x^2-1$
$ = -(x^2-4x+4) - 1 + 4$
$ = -(x-2)^2+3 ≤ 3$
Dấu "=" xảy ra $⇔x=2$
f) $H = 5-x^2+2x-4y^2-4y$
$ = -(x^2-2x+1) - (4y^2+4y+1) + 7$
$ = -(x-1)^2-(2y-1)^2+7 ≤ 7$
Dấu "=" xảy ra $⇔x=1,y=\dfrac{1}{2}$
Tính $GTNN$ :
$A = 5x^2+5y^2+8xy+2y-2x+2020$
$ \to 5A = 25x^2+25y^2+40xy+10y-10x+10100$
$ =(5x)^2 + 2.5x.4y + (4y)^2 + 9y^2+10y-10x + 10100$
$ = (5x+4y)^2 - 2.(5x+4y).1 + 1 + 9y^2+18y + 10099$
$ = (5x+4y-1)^2+9.(y^2+2y+1) + 10090$
$ = (5x+4y-1)^2 + 9.(y+1)^2 + 10090 ≥ 10090$
$\to A ≥ \dfrac{10090}{4} = \dfrac{5045}{2}$
Dấu "=" xảy ra $⇔x=1,y=-1$
Vậy $A_{min} = \dfrac{5045}{2}$ khi $x=1,y=-1$
