a, $A=x^2-x+1$
$=(x^2-x+1/4)+3/4$
$=(x-1/2)^2+3/4≥3/4$
Daaus "=" xayr ra khi $x-1/2=0⇒x=1/2$
Vaayj $minA=3/4$ khi $x=1/2$
b, $A=(x-1)(x-2)(x-3)(x-4)+15$
$=(x^2-5x+4)(x^2-5x+6)+15$
Đặt: $y=x^2-5x+5$
$⇒A=(y-1)(y+1)+15$
$=y^2-1+15$
$=(x^2-5x+5)^2+14≥14$
Dấu "=" xảy ra khi $x^2-5x+5=0⇒$$x=\frac{5±\sqrt[]{5}}{2}$
Vậy $minC=14$ khi $x=\frac{5±\sqrt[]{5}}{2}$
c, $1-x^2-x^4≤1$
Dấu "=" xảy ra khi $x=0$
Vậy $maxC=1$ khi $x=0$