12. 计算：
(1) $\sqrt{\dfrac{1}{2}} × \sqrt{32} + \dfrac{\sqrt{18} - \sqrt{8}}{\sqrt{2}}$；
(2) $\sqrt{48} ÷ \sqrt{3} + 2\sqrt{\dfrac{1}{5}} × \sqrt{30} - (2\sqrt{2} + \sqrt{3})^2$。

(1) 5 (2) $-7-2\sqrt{6}$

13. 已知 $x=\sqrt{5}+\sqrt{3}$，$y=\sqrt{5}-\sqrt{3}$，求 $x^2 + 3xy + y^2$ 的值。

13. $\because x=\sqrt{5}+\sqrt{3},\ y=\sqrt{5}-\sqrt{3},\therefore x+y=(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=2\sqrt{5},\ xy=(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=2.$$\therefore x^2+3xy+y^2=x^2+2xy+y^2+xy=(x+y)^2+xy=(2\sqrt{5})^2+2=22$

14. 阅读下列材料，然后解答问题.
$\frac{1}{\sqrt{2}+1}=\frac{1×(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1×(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$；
$\frac{1}{\sqrt{5}+2}=\frac{1×(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}=\sqrt{5}-2$.
以上这种化简的方法叫做分母有理化.
(1) $\frac{1}{\sqrt{7}+\sqrt{6}}=$ .
(2) $\frac{1}{\sqrt{n+1}+\sqrt{n}}$（$n$为正整数）= .
(3) 化简：$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\dots+\frac{1}{\sqrt{100}+\sqrt{99}}=$ .
(4) 化简：$\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\dots+\frac{1}{\sqrt{99}+\sqrt{97}}$.

14. (1) $\sqrt{7}-\sqrt{6}$ (2) $\sqrt{n+1}-\sqrt{n}$ (3) 9 (4) 原式$=\frac{1}{2}(\sqrt{3}-1)+\frac{1}{2}(\sqrt{5}-\sqrt{3})+\frac{1}{2}(\sqrt{7}-\sqrt{5})+\dots+\frac{1}{2}(\sqrt{99}-\sqrt{97})=\frac{1}{2}(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\dots+\sqrt{99}-\sqrt{97})=\frac{1}{2}(\sqrt{99}-1)=\frac{3\sqrt{11}-1}{2}$