A= x2-20x+101
= x2-20x+100+1
= (x2-20x+100)+1
= (x-10)2+1
do (x-10)2 ≥ 0 ∀ x
⇔ (x-10)2+1 ≥ 1 ∀ x
⇔ A ≥ 1 ∀ x
=> min A =1 khi x=10
B= x2-4xy+5y2+10x-22y+28
= (x2-4xy+4y2)+ (10x+20y) +25+(y2+2y+1)+2
= [(x-2y)2+10(x-2y)+25]+(y+1)2+2
= (x-2y+5)2+(y+1)2+2
do (x-2y+5)2 ≥ 0∀ x;y
(y+1)2 ≥ 0∀ y
=> (x-2y+5)2 + (y+1)2 ≥ 0∀ x;y
⇔ (x-2y+5)2+(y+1)2+2 ≥ 2∀ x;y
⇔ B ≥ 2∀ x;y
min B =2 khi y=-1;x=-3
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