12.我们规定运算符号“△”：当$a ＞ b$时，$a△b = a + b$；当$a ≤ b$时，$a△b = a - b$，其他运算符号的意义不变.计算：$(\sqrt{3}△\sqrt{2}) - (2\sqrt{3}△3\sqrt{2}) = \_\_\_\_\_\_.$

12.4√2−√3

三、解答题
13. 计算：
(1) $\sqrt{12} - \sqrt{18} + 3\sqrt{\frac{1}{3}} + \sqrt{8}$；
(2) $(\sqrt{5} + 1)(\sqrt{5} - 1) - (\sqrt{3} - \sqrt{2})^2$。

13.(1)3√3−√2 (2)2√6−1

14. 在进行二次根式化简时，我们有时会碰见如$\frac{5}{\sqrt{3}},\sqrt{\frac{2}{3}},\frac{2}{\sqrt{3}+1}$一类的式子，其实我们还可以将其进一步化简：$\frac{5}{\sqrt{3}}=\frac{5×\sqrt{3}}{\sqrt{3}×\sqrt{3}}=\frac{5\sqrt{3}}{3}$；$\sqrt{\frac{2}{3}}=\sqrt{\frac{2×3}{3×3}}=\frac{\sqrt{6}}{3}$；$\frac{2}{\sqrt{3}+1}=\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\sqrt{3}-1$。以上这种化简的步骤叫做分母有理化。$\frac{2}{\sqrt{3}+1}$还可以用以下方法化简：

(1)请用两种不同的方法化简$\frac{2}{\sqrt{5}+\sqrt{3}}$；
(2)化简：$\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\dots+\frac{1}{\sqrt{21}+\sqrt{19}}$

14.解:(1)方法一:$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\sqrt{5}-\sqrt{3}$.
方法二:$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5})^2-(\sqrt{3})^2}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}{\sqrt{5}+\sqrt{3}}=\sqrt{5}-\sqrt{3}$.
(2)$\because \frac{1}{\sqrt{2n+1}+\sqrt{2n-1}}$
$=\frac{1}{2}·\frac{(2n+1)-(2n-1)}{\sqrt{2n+1}+\sqrt{2n-1}}$
$=\frac{1}{2}·\frac{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n+1}-\sqrt{2n-1})}{\sqrt{2n+1}+\sqrt{2n-1}}$
$=\frac{\sqrt{2n+1}-\sqrt{2n-1}}{2}$(n为正整数),
$\therefore \frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\dots+\frac{1}{\sqrt{21}+\sqrt{19}}$
$=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\dots+\frac{\sqrt{21}-\sqrt{19}}{2}$
$=\frac{\sqrt{21}-1}{2}$.