4. 已知$\sqrt{a + 2}+\vert b - 1\vert=0$，那么$(a + b)^{2015}=$_______.

$-1$

5. 在$\triangle ABC$中，$AB = 2\sqrt{2}$，$BC = 1$，$\angle ABC = 45^{\circ}$，以$AB$为一边作等腰直角三角形$ABD$，使$\angle ABD = 90^{\circ}$，连接$CD$，则线段$CD$的长为_______.

$\sqrt{5}$ 或 $\sqrt{13}$

6. 先化简，再求值：$(\frac{1}{a - 2}-\frac{3}{a^{2}-4})\div\frac{a - 1}{a^{2}+2a}$，其中$a = \sqrt{6}+2$.

$\frac{a}{a - 2}$，$\frac{3 + \sqrt{6}}{3}$

7. 比较$\sqrt{2}-1$与$\sqrt{3}-\sqrt{2}$的大小可以采用下面的方法：
$\sqrt{2}-1=\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}$
$=\frac{2 - 1}{\sqrt{2}+1}$
$=\frac{1}{\sqrt{2}+1}$，
$\sqrt{3}-\sqrt{2}=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}$
$=\frac{3 - 2}{\sqrt{3}+\sqrt{2}}$
$=\frac{1}{\sqrt{3}+\sqrt{2}}$.
显然$\sqrt{2}+1＜\sqrt{3}+\sqrt{2}$，所以$\frac{1}{\sqrt{2}+1}＞\frac{1}{\sqrt{3}+\sqrt{2}}$.
仔细研读上面的解题方法，然后完成下列问题：
(1)猜想：$\sqrt{2011}-\sqrt{2010}$与$\sqrt{2012}-\sqrt{2011}$的大小关系；
(2)尝试计算：$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{2+\sqrt{3}}+\cdots+\frac{1}{10+\sqrt{99}}$.

(1) $\sqrt{2011} - \sqrt{2010} = \frac{(\sqrt{2011} - \sqrt{2010})(\sqrt{2011} + \sqrt{2010})}{1 \times (\sqrt{2011} + \sqrt{2010})} = \frac{1}{\sqrt{2011} + \sqrt{2010}}$，$\sqrt{2012} - \sqrt{2011} = \frac{(\sqrt{2012} - \sqrt{2011})(\sqrt{2012} + \sqrt{2011})}{(\sqrt{2012} + \sqrt{2011}) \times 1} = \frac{1}{\sqrt{2012} + \sqrt{2011}}$. 显然 $\sqrt{2011} + \sqrt{2010} ＜ \sqrt{2012} + \sqrt{2011}$，所以 $\frac{1}{\sqrt{2011} + \sqrt{2010}} ＞ \frac{1}{\sqrt{2012} + \sqrt{2011}}$. 所以 $\sqrt{2011} - \sqrt{2010} ＞ \sqrt{2012} - \sqrt{2011}$
(2) $\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{2 + \sqrt{3}} + \cdots + \frac{1}{10 + \sqrt{99}} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} + \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} + \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} + \cdots + \frac{10 - \sqrt{99}}{(10 + \sqrt{99})(10 - \sqrt{99})} = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + \cdots + (10 - \sqrt{99}) = -1 + 10 = 9$