2,
a) A = x² - 6x + 11 = x² - 6x + 9 + 2 = ( x - 3 )² + 2 ≥ 2
Dấu " = " xảy ra khi x - 3 = 0 ⇔ x = 3
Vậy Min A = 2 khi x = 3
b) B = x² - 20x + 101 = x² - 20x + 100 + 1 = ( x - 10 )² + 1 ≥ 1
Dấu " = " xảy ra khi x - 10 = 0 ⇔ x = 10
Vậy Min B = 1 khi x = 10
c) C = x² - 4xy + 5y² + 10x - 22y + 28
= ( x² + 4y² + 25 - 4xy + 10x - 20y ) + ( y² - 2y + 1 ) + 2
= ( x - 2y + 5 )² + ( y - 1 )² + 2 ≥ 2
Dấu " = " xảy ra khi $\left \{ {{x-2y+5=0} \atop {y-1=0}} \right.$ ⇔ $\left \{ {{x=-3} \atop {y=1}} \right.$
Vậy Min C = 2 khi $\left \{ {{x=-3} \atop {y=1}} \right.$
3,
a) A = 5x - x² = - ( x² - 2.x.$\frac{5}{2}$ + $\frac{25}{4}$ - $\frac{25}{4}$ ) = $\frac{25}{4}$ - ( x - $\frac{5}{2}$ )² ≤ $\frac{25}{4}$
Dấu " = " xảy ra khi x - $\frac{5}{2}$ = 0 ⇔ x = $\frac{5}{2}$
Vậy Max A = $\frac{25}{4}$ khi x = $\frac{5}{2}$
b) B = x - x² = - ( x² - 2.x.$\frac{1}{2}$ + $\frac{1}{4}$ - $\frac{1}{4}$ ) = $\frac{1}{4}$ - ( x - $\frac{1}{2}$ )² ≤ $\frac{1}{4}$
Dấu " = " xảy ra khi x - $\frac{1}{2}$ = 0 ⇔ x = $\frac{1}{2}$
Vậy Max B = $\frac{1}{4}$ khi x = $\frac{1}{2}$
c) C = 4x - x² + 3 = - ( x² - 4x - 3 ) = - ( x² - 4x + 4 - 7 ) = 7 - ( x - 2 )² ≤ 7
Dấu " = " xảy ra khi x - 2 = 0 ⇔ x = 2
Vậy Max C = 7 khi x = 2
