1/ Thay $x=-2$ vào biểu thức $A$:
$A\,=3.(-2)^2-4.(-2)+1\\\quad =3.4+8+1\\\quad =21$
Vậy $A=21$ tại $x=-2$
2/ $B\,=(x-1)(x^2+x+1)-(x+1)(2-x)-x^3\\\quad =x^3-1-[x(2-x)+1(2-x)]-x^3\\\quad =(x^3-x^3)-1-[2x-x^2+2-x]\\\quad =-1-[-x^2+x+2]\\\quad =-1+x^2-x-2\\\quad =x^2-x-3$
Vậy $B=x^2-x-3$
3/ $C\,=B-A\\\quad =(x^2-x-3)-(3x^2-4x+1)\\\quad =x^2-x-3-3x^2+4x-1\\\quad =-2x^2+3x-4\\\quad =-2\left(x^2-\dfrac{3}{2}x+2\right)\\\quad =-2\left(x^2-2.x.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}\right)\\\quad =-2\left(x-\dfrac{3}{4}\right)^2-\dfrac{23}{8}$
Nhận thấy: $\left(x-\dfrac{3}{4}\right)^2\ge 0$
$↔-2\left(x-\dfrac{3}{4}\right)^2\le 0\\↔C\le -\dfrac{23}{8}\\→\max C=-\dfrac{23}{8}$
$→$ Dấu "=" xảy ra khi $x-\dfrac{3}{4}=0$
$↔x=\dfrac{3}{4}$
Vậy $\max C=-\dfrac{23}{8}$ khi $x=\dfrac{3}{4}$
