18.若$0＜x＜1$，试化简：$|x|+(\sqrt{1-x})^2-\sqrt{(x-1)^2}-\sqrt{(x-2)^0}$。

18.$\because 0＜x＜1$,
$\therefore$原式$=x+1-x-|x-1|-1=x-1$.

19. 已知 $ x - 1 = \sqrt{3} $，求代数式 $ (x + 1)^2 - 4(x + 1) + 4 $ 的值.

19.$(x+1)^2-4(x+1)+4$
$=[(x+1)-2]^2$
$=(x-1)^2$.
$\because x-1=\sqrt{3}$,
$\therefore$原式$=(\sqrt{3})^2$
$=3$.

20.阅读下列解题过程：

请回答下列问题：
(1)仿照上面的解题过程化简：$\frac{1}{\sqrt{6}+\sqrt{5}}=\_\_\_\_\_\_=\_\_\_\_\_\_=$.
(2)化简：$\frac{1}{\sqrt{n+1}+\sqrt{n}}=$.
(3)利用上面所提供的方法，求
$( \frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}} + \dots + \frac{1}{\sqrt{2025}+\sqrt{2024}} ) × (\sqrt{2025} + 1)$
的值.
(4)利用上面的结论，不计算近似值，试比较$(\sqrt{12}-\sqrt{11})$与$(\sqrt{13}-\sqrt{12})$的大小.

20.(1)$\frac{1×(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})}$;$\frac{\sqrt{6}-\sqrt{5}}{(\sqrt{6})^2-(\sqrt{5})^2}$;$\sqrt{6}-\sqrt{5}$
(2)$\sqrt{n+1}-\sqrt{n}$
(3)$[\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}+\dots+\frac{\sqrt{2025}-\sqrt{2024}}{(\sqrt{2025}+\sqrt{2024})(\sqrt{2025}-\sqrt{2024})}]×(\sqrt{2025}+1)$
$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\dots+\sqrt{2025}-\sqrt{2024})(\sqrt{2025}+1)$
$=(\sqrt{2025}-1)(\sqrt{2025}+1)=2025-1$
$=2024$.
(4)利用上述结论可得$\frac{1}{\sqrt{12}+\sqrt{11}}=\sqrt{12}-\sqrt{11}$,$\frac{1}{\sqrt{13}+\sqrt{12}}=\sqrt{13}-\sqrt{12}$.
$\because \sqrt{12}+\sqrt{11}＜\sqrt{13}+\sqrt{12}$,
$\therefore \frac{1}{\sqrt{12}+\sqrt{11}}＞\frac{1}{\sqrt{13}+\sqrt{12}}$.
$\therefore \sqrt{12}-\sqrt{11}＞\sqrt{13}-\sqrt{12}$.