24. (11 分)如图1,在平面直角坐标系中,点$A(a,0)$,点$B(0,b)$,且$a,b$满足$|a + 3| + \sqrt{b + 4} = 0$.
(1)求点$A,B$的坐标;
(2)将线段$AB$平移至$CD$处,点$C$与点$A$是对应点,$C(x,y)$,$F$是线段$CD$上的点,点$F$的坐标为$(x + 1.5,y - 2)$,$E$是$x$轴上一动点,点$E$的坐标为$(m,0)$.
①当$x = \frac{9}{2},y = 6$,且$m ＞ 3$时,$△ CFE$的面积与$△ BFE$的面积相等,求$m$的值;
②如图2,若点$C,D$均在第一象限,且满足$4x + 12 = ny$,连接$AC,BD,ED$,若$△ ACE$的面积小于$△ BED$的面积,且不小于$△ BED$面积的一半,直接写出$m$的取值范围:________. (用含$n$的式子表示)

24.
【解析】(1)
∵ $|a + 3| + \sqrt{b + 4} = 0$,
∴ $a + 3 = 0$,$b + 4 = 0$,
∴ $a = - 3$,$b = - 4$,
∴ $A(- 3,0)$,$B(0,- 4)$.
(2)①如图1,连接$OF$,设$BF$交$x$轴于点$G$,延长$CD$交$x$轴于点$H$,过点$D$作$DN⊥x$轴于点$N$,过点$C$作$CM⊥x$轴于点$M$.
∵ $x = \dfrac{9}{2}$ ,$y = 6$,
∴ $C(\dfrac{9}{2} ,6)$,$F(6,4)$,
∴ $CM = 6$,$OM = \dfrac{9}{2}$ . 由平移的性质可得$D(\dfrac{15}{2} ,2)$,
∴ $DN = 2$,$ON = \dfrac{15}{2}$ ,
∴ $MN = 3$. 设$NH = p$,则$MH = p + 3$.
∵ $S_{△ CMH} = S_{△ DNH} + S_{梯形CMND}$,
∴ $\dfrac{1}{2} ×6×(p + 3) = \dfrac{1}{2} ×2×p + \dfrac{1}{2} ×(2 + 6)×3$,解得$p = \dfrac{3}{2}$ ,即$NH = \dfrac{3}{2}$ ,
∴ $OH = ON + NH = 9$,
∴ $H(9,0)$.
∵ $S_{△ OBF} = \dfrac{1}{2} OB · x_F = \dfrac{1}{2} ×4×6 = 12$,
∴ $S_{△ OBF} = S_{△ OBG} + S_{△ OFG} = \dfrac{1}{2} OG · OB + \dfrac{1}{2} OG · y_F = 4OG$,
∴ $4OG = 12$,
∴ $OG = 3$,
∴ $G(3,0)$.
∵ $E(m,0)$,$m ＞ 3$,
∴ $EG = m - 3$,$EH = |m - 9|$,
∴ $S_{△ BFE} = S_{△ BEG} + S_{△ EFG} = \dfrac{1}{2} (m - 3)×4 + \dfrac{1}{2} (m - 3)×4 = 4(m - 3)$.
∵ $S_{△ CFE} = S_{△ CEH} - S_{△ FEH} = \dfrac{1}{2} |m - 9|×6 - \dfrac{1}{2} |m - 9|×4 = |m - 9|$,
$S_{△ BFE} = S_{△ CFE}$,
∴ $4(m - 3) = |m - 9|$,解得$m = \dfrac{21}{5}$ 或$m = 1$(舍去),
∴ $m$的值为$\dfrac{21}{5}$.
②如图2,过点$D$作$DG⊥x$轴于点$G$,过点$B$作$BH⊥DG$交$DG$的延长线于点$H$,设$BD$交$x$轴于点$K$,连接$KH$.
已知$C(x,y)$,$E(m,0)$,由平移性质可得$D(x + 3,y - 4)$.
设$K(k,0)$,则$KG = x + 3 - k$.
∵ $S_{△ BDH} = S_{△ DKH} + S_{△ BHK}$,
∴ $\dfrac{1}{2} (x + 3)y = \dfrac{1}{2} (x + 3 - k)y + \dfrac{1}{2} (x + 3)×4$,整理得$ky = 4x + 12$,
∴ $k = \dfrac{4x + 12}{y} = n$,
∴ $K(n,0)$.
当点$E$在点$A,K$之间时,$n ＞ m$,
∴ $S_{△ ACE} = \dfrac{1}{2} (m + 3)y$,$EK = n - m$,
∴ $S_{△ BED} = S_{△ BEK} + S_{△ DEK} = \dfrac{1}{2} (n - m)y$.
∵ $S_{△ ACE} ＜ S_{△ BED}$ ,$S_{△ ACE} ≥ \dfrac{1}{2} S_{△ BED}$,
∴ $\begin{cases}\dfrac{1}{2}(m + 3)y ＜ \dfrac{1}{2}(n - m)y, \\\dfrac{1}{2}(m + 3)y ≥ \dfrac{1}{2}×\dfrac{1}{2}(n - m)y,\end{cases}$解得$\dfrac{n - 6}{3} ≤ m ＜ \dfrac{n - 3}{2}$.
当点$E$在点$A$左侧时,$S_{△ ACE} = \dfrac{-3 - m}{2} · y$,$S_{△ BED} = \dfrac{1}{2} (n - m)y$,
∵ $\dfrac{1}{2} S_{△ BED} ≤ S_{△ ACE} ＜ S_{△ BED}$,
∴ $n - m ≤ 2(- 3 - m) ＜ 2(n - m)$,
∴ $m ≤ - n - 6$.
当点$E$在点$K$右侧时,$S_{△ AEC} ＞ S_{△ BED}$,不符合题意.
综上所述,$m$的取值范围为$\dfrac{n - 6}{3} ≤ m ＜ \dfrac{n - 3}{2}$ 或$m ≤ - n - 6$. 故答案为$\dfrac{n - 6}{3} ≤ m ＜ \dfrac{n - 3}{2}$ 或$m ≤ - n - 6$.