$C=x^2-4xy+5y^2+10x-22y$
$=[(x^2-4xy+4y^2)+10.(x-2y)+25]+(y^2-2y+1)-26$
$=(x-2y+5)^2+(y-1)^2-26≥-26$
Dấu "=" xảy ra khi $\left \{ {{x-2y+5=0} \atop {y-1=0}} \right.$ $⇒\left \{ {{x=-3} \atop {y=1}} \right.$
Vậy $minC=-26$ khi $\left \{ {{x=-3} \atop {y=1}} \right.$