三、解答题(共48分)
8. (24分)计算:
(1) $\sqrt{18} - 4\sqrt{\frac{1}{2}} + \sqrt{24} ÷ \sqrt{3}$；
(2) $\frac{1}{6}\sqrt{1\frac{3}{5}} × (-5\sqrt{3\frac{3}{5}}) ÷ (-2\sqrt{\frac{1}{2}})$；
(3) $\sqrt{42} ÷ \sqrt{6} - \sqrt{\frac{1}{3}} × \sqrt{15} - (\sqrt{7} - 3)$；
(4) $\frac{2}{\sqrt{3} - 2} - (3 - 2\sqrt{2})(3 + 2\sqrt{2})$。

8. (1) $3\sqrt{2}$ (2) $\sqrt{2}$ (3) $3-\sqrt{5}$ (4) $-2\sqrt{3}-5$

9. (10分)先化简,再求值: $(\dfrac{a}{a^2 -4} + \dfrac{1}{2 -a}) ÷ \dfrac{2a +4}{a^2 +4a +4}$,其中 $a=\sqrt{3}+2$.

9. 原式$=\dfrac{1}{2-a}$. 当 $a=\sqrt{3}+2$ 时,原式$=\dfrac{1}{2-\sqrt{3}-2}=-\dfrac{\sqrt{3}}{3}$

10. (14分)[贵州中考]如图,在$□ ABCD$中,E为对角线AC的中点,连接BE,且$BE ⊥ AC$,延长BC至点F,使$CF=CE$,连接EF,FD,且EF交CD于点G.
(1) 求证:$□ ABCD$是菱形;
(2) 若$BE=EF,EC=4$,求$△ DCF$的面积.

10. (1) $\because$ E为对角线AC的中点,$BE ⊥ AC$,$\therefore$ BE垂直平分AC. $\therefore AB=BC$. $\because$ 四边形ABCD是平行四边形, $\therefore □ ABCD$ 是菱形
(2) $\because BE = EF$,$\therefore ∠EBF=∠EFB$. $\because CF=CE$,$\therefore ∠CEF=∠CFE$.
$\therefore ∠BCE = ∠CEF + ∠CFE = 2 ∠CFE = 2 ∠EBF$.
$\because ∠BEC=90°$,$\therefore$ 易得$∠CBE=30°$,$∠BCA=60°$.
在菱形ABCD中,$BC = CD$,$∠ACB = ∠ACD = 60°$.
$\therefore ∠DCF = 60°$. $\therefore ∠BCE = ∠DCF$. $\because BC = DC$,
$\therefore △ BCE ≌ △ DCF$. $\therefore ∠BEC = ∠DFC = 90°$.
$\because CF=CE=4$,$\therefore$ 易得 $DF=4\sqrt{3}$. $\therefore △ DCF$ 的面积$=\dfrac{1}{2}DF· CF=8\sqrt{3}$