Đáp án:
$g)x^2-4xy+5y^2+10x-22y+28$
$=x^2-2x(2y-5)+(4y^2-20y+25)+y^2-2y+3$
$=x^2-2x(2y-5)+(2y-5)^2+y^2-2y+1+2$
$=(x-2y+5)^2+(y-1)^2+2≥2∀x;y$
Dấu "=" xảy ra ⇔$\left \{ {{x-2y+5=0} \\ {y-1=0}} \right.$
⇔$\left \{ {{y=1} \\ {x=-3}} \right.$
Vậy ...................