21. (12分)定义:如果两个多项式M与N的和为常数,则称M与N互为“对消多项式”,这个常数称为它们的“对消值”.如$M=2x^2 - x + 6$与$N = -2x^2 + x - 1$互为“对消多项式”,它们的“对消值”为5.
(1)下列各组多项式互为“对消多项式”的是________;(填序号)
①$3x^2 + 2x$与$3x^2 + 2$;②$x - 6$与$-x + 2$;③$-5x^2y^3 + 2xy$与$5x^2y^3 - 2xy - 1$.
(2)多项式$A=(x - a)^2$与多项式$B = -bx^2 - 2x + b$($a,b$为常数)互为“对消多项式”,求它们的“对消值”;
(3)关于$x$的多项式$C = mx^2 + 6x + 4$与$D = -m(x + 1)(x + n)$互为“对消多项式”,“对消值”为$t$.若$a - b = m,b - c = mn$,求代数式$a^2 + b^2 + c^2 - ab - bc - ac + 2t$的最小值.

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【解析】(1)$\because 3x^2 +2x +3x^2 +2 =6x^2 +2x +2$,$x-6 -x +2 =-4$,$-5x^2y^3 +2xy +5x^2y^3 -2xy -1 =-1$,$\therefore$ ①组多项式不互为“对消多项式”,②③组多项式互为“对消多项式”.
故答案为②③.
(2)$\because A=(x-a)^2 =x^2 -2ax +a^2$,$B=-bx^2 -2x +b$,
$\therefore A+B=(1-b)x^2 +(-2a -2)x +(a^2 +b)$.
$\because A$与$B$互为“对消多项式”,
$\therefore 1-b=0$,$-2a-2=0$,$\therefore a=-1$,$b=1$,
$\therefore$ 它们的“对消值”为$a^2 +b =2$.
(3)$\because C=mx^2 +6x +4$,$D=-m(x+1)(x+n) =-mx^2 +(-mn -m)x -mn$,$\therefore C+D=(6 -mn -m)x +(4 -mn)$.
$\because C$与$D$互为“对消多项式”且“对消值”为$t$,
$\therefore \begin{cases}6 - mn -m =0,\\4 - mn =t,\end{cases}$ $\therefore \begin{cases}mn=6 -m,\\t=4 -mn =4 -(6 -m) =m -2.\end{cases}$
$\because a -b =m$,$b -c =mn$,
$\therefore a -c =(a -b)+(b -c) =m +mn =6$,
$\therefore a^2 +b^2 +c^2 -ab -bc -ac +2t$
$=\dfrac{1}{2}[(a-b)^2 +(b-c)^2 +(a-c)^2] +2t$
$=\dfrac{1}{2}[m^2 +(mn)^2 +6^2] +2(m-2)$
$=\dfrac{1}{2}[m^2 +(6 -m)^2 +6^2] +2(m-2)$
$=m^2 -4m +32$
$=(m-2)^2 +28$.
$\because (m-2)^2 \ge0$,$\therefore (m-2)^2 +28 \ge28$,
$\therefore$ 代数式$a^2 +b^2 +c^2 -ab -bc -ac +2t$的最小值是28.