20. (6分)先化简,再求值:$(x+2y)(x-2y)+(x-2y)^2-(6x^2y-2xy^2)÷2y$,其中$x=-3,y=\dfrac{1}{3}$.

【点拨】本题考查整式的混合运算.
【解析】 $(x+2y)(x-2y)+(x-2y)^2-(6x^2y-2xy^2)÷2y$
$=x^2-4y^2 +x^2-4xy +4y^2 -3x^2 +xy$
$=-x^2 -3xy$,
当$x=-3,y=\dfrac{1}{3}$时,
原式$=-(-3)^2 -3×(-3)×\dfrac{1}{3}$
$=-9+3$
$=-6$.

21. (8 分)(1)解方程组: $\begin{cases}3x + z = 9, \\x + 2y - z = 3, \\x + y + z = 7;\end{cases}$ (2)解不等式组: $\begin{cases}5x - 3 ≤ 2x + 9, \\3x ＞ \dfrac{x + 10}{2}.\end{cases}$

【解析】(1)$\begin{cases}3x + z = 9,① \\x + 2y - z = 3,② \\x + y + z = 7,③\end{cases}$
① + ②,得$4x + 2y = 12$,
$\therefore 2x + y =6$,④
① - ③,得$2x - y =2$.⑤
④ + ⑤,得$4x=8$. 解得$x=2$.
将$x=2$,代入④,得$y=2$,代入①,得$z=3$.
$\therefore$ 原方程组的解为$\begin{cases} x=2, \\ y=2, \\ z=3. \end{cases}$
(2)$\begin{cases}5x -3 ≤ 2x +9,① \\3x ＞ \dfrac{x+10}{2},②\end{cases}$
解不等式①,得$x≤4$.
解不等式②,得$x＞2$.
$\therefore$ 不等式组的解集为$2 ＜ x≤4$.

22. (6分)如图,在$△ ABC$中,边$BC$的垂直平分线$MN$分别交边$AC$于点$D$,边$BC$于点$E$,连接$BD$.
(1)若$CE=4$,$△ BDC$的周长为$18$,求$BD$的长;
(2)若$∠ ADM=60°$,$∠ ABD=20°$,求$∠ A$的度数.

【点拨】本题考查垂直平分线的性质,等腰三角形三线合一的性质及角的和差.
【解析】(1)$\because MN$垂直平分$BC$,$CE=4$,
$\therefore DC=DB$,$BE=CE=4$,
$\therefore BC=BE+CE=4+4=8$.
$\because △ BDC$的周长为18,
$\therefore BD+DC+BC=BD+DC+8=18$,
$\therefore BD=DC=5$.
(2)$\because ∠ ADM=60°$,$∠ ABD=20°$,$\therefore ∠ CDN=∠ ADM=60°$.
$\because MN$垂直平分$BC$,$\therefore DC=DB$,$MN⊥ BC$,
$\therefore ∠ BDN=∠ CDN=60°$,
$\therefore ∠ BDC=∠ BDN+∠ CDN=60°+60°=120°$,
$\therefore ∠ A=∠ BDC-∠ ABD=120°-20°=100°$,
$\therefore ∠ A$的度数为$100°$.