一个梯形的上、下底的长分别为 $ 2\sqrt{2} $，$ 4\sqrt{3} $，高为 $ \sqrt{6} $，你能求出它的面积吗？写出结果并化简.

梯形面积公式：$ S = \frac{1}{2}(a + b)h $，其中$ a = 2\sqrt{2} $，$ b = 4\sqrt{3} $，$ h = \sqrt{6} $。
$ S = \frac{1}{2}(2\sqrt{2} + 4\sqrt{3})×\sqrt{6} $
$ = \frac{1}{2}×2\sqrt{2}×\sqrt{6} + \frac{1}{2}×4\sqrt{3}×\sqrt{6} $
$ = \sqrt{12} + 2\sqrt{18} $
$ = 2\sqrt{3} + 2×3\sqrt{2} $
$ = 2\sqrt{3} + 6\sqrt{2} $
结果：$ 6\sqrt{2} + 2\sqrt{3} $

例 计算：
(1) $ (\sqrt{2}-1)^2 × (3 + 2\sqrt{2}) $；
(2) $ (\sqrt{3}+1-\sqrt{2}) × (\sqrt{3}+1+\sqrt{2}) $.

(1)
$(\sqrt{2}-1)^2×(3 + 2\sqrt{2})$
$=( (\sqrt{2})^2 - 2×\sqrt{2}×1 + 1^2 )×(3 + 2\sqrt{2})$
$=(2 - 2\sqrt{2} + 1)×(3 + 2\sqrt{2})$
$=(3 - 2\sqrt{2})(3 + 2\sqrt{2})$
$=3^2 - (2\sqrt{2})^2$
$=9 - 8$
$=1$
(2)
$(\sqrt{3}+1-\sqrt{2})(\sqrt{3}+1+\sqrt{2})$
$=[(\sqrt{3}+1)-\sqrt{2}][(\sqrt{3}+1)+\sqrt{2}]$
$=(\sqrt{3}+1)^2 - (\sqrt{2})^2$
$=( (\sqrt{3})^2 + 2×\sqrt{3}×1 + 1^2 ) - 2$
$=(3 + 2\sqrt{3} + 1) - 2$
$=4 + 2\sqrt{3} - 2$
$=2 + 2\sqrt{3}$

1. 计算：
(1) $ ( \sqrt{6}-\sqrt{\dfrac{3}{8}} ) × \sqrt{2} $；
(2) $ (2+\sqrt{2}) × (1-\sqrt{2}) $；
(3) $ (\sqrt{2}+1) × (\sqrt{2}-1) $；
(4) $ (\sqrt{2}-\sqrt{3})^2 $.

(1)
$\begin{aligned}&(\sqrt{6}-\sqrt{\dfrac{3}{8}})×\sqrt{2}\\=&\sqrt{6}×\sqrt{2}-\sqrt{\dfrac{3}{8}}×\sqrt{2}\\=&\sqrt{12}-\sqrt{\dfrac{3}{8}×2}\\=&\sqrt{12}-\sqrt{\dfrac{3}{4}}\\ = &2\sqrt{3}-\dfrac{\sqrt{3}}{2}\\=&\dfrac{3\sqrt{3}}{2}\end{aligned}$
(2)
$\begin{aligned}&(2+\sqrt{2})×(1-\sqrt{2})\\ =&2×1 - 2×\sqrt{2}+\sqrt{2}×1-\sqrt{2}×\sqrt{2}\\=&2 - 2\sqrt{2}+\sqrt{2}- 2\\ =&-\sqrt{2}\end{aligned}$
(3)
$\begin{aligned}&(\sqrt{2}+1)×(\sqrt{2}-1)\\ =&(\sqrt{2})^{2}-1^{2}\\ =&2 - 1\\ =&1\end{aligned}$
(4)
$\begin{aligned}&(\sqrt{2}-\sqrt{3})^{2}\\ =&(\sqrt{2})^{2}-2×\sqrt{2}×\sqrt{3}+(\sqrt{3})^{2}\\ =&2-2\sqrt{6}+3\\ =&5 - 2\sqrt{6}\end{aligned}$

2. 计算：
(1) $ a\sqrt{16a}+3\sqrt{a^3}-\dfrac{1}{2}a^2\sqrt{\dfrac{4}{a}}(a＞0) $；
(2) $ (2+\sqrt{5}) × (3-2\sqrt{5}) $；
(3) $ (3\sqrt{2}+\sqrt{7}) × (3\sqrt{2}-\sqrt{7}) $；
(4) $ (\sqrt{5}-\sqrt{10})^2 $.

(1)
$\because a＞0$，
$\therefore a\sqrt{16a}+3\sqrt{a^3}-\dfrac{1}{2}a^2\sqrt{\dfrac{4}{a}}$
$=4a\sqrt{a}+3a\sqrt{a}-\dfrac{1}{2}a^2×\dfrac{2\sqrt{a}}{a}$
$=4a\sqrt{a}+3a\sqrt{a}-a\sqrt{a}$
$=(4a + 3a - a)\sqrt{a}$
$=6a\sqrt{a}$
(2)
$(2+\sqrt{5})×(3 - 2\sqrt{5})$
$=2×3-2×2\sqrt{5}+3\sqrt{5}-2×5$
$=6 - 4\sqrt{5}+3\sqrt{5}-10$
$=(6 - 10)+(-4\sqrt{5}+3\sqrt{5})$
$=-4-\sqrt{5}$
(3)
$(3\sqrt{2}+\sqrt{7})×(3\sqrt{2}-\sqrt{7})$
$=(3\sqrt{2})^2-(\sqrt{7})^2$
$=18 - 7$
$=11$
(4)
$(\sqrt{5}-\sqrt{10})^2$
$=(\sqrt{5})^2-2×\sqrt{5}×\sqrt{10}+(\sqrt{10})^2$
$=5 - 2\sqrt{50}+10$
$=5 - 10\sqrt{2}+10$
$=15 - 10\sqrt{2}$