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11. 阅读下面材料，并解决相应的问题：
$\dfrac{3}{\sqrt{5}-\sqrt{2}}=\dfrac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}=\dfrac{3(\sqrt{5}+\sqrt{2})}{3}=\sqrt{5}+\sqrt{2}$.
(1)用上述类似的方法化简下列各式：
①$\dfrac{1}{\sqrt{3}+\sqrt{2}}=$
②计算：$\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dots+\dfrac{1}{\sqrt{99}+\sqrt{100}}$；
③计算：$\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dots+\dfrac{1}{\sqrt{99}-\sqrt{100}}$.
(2)已知实数$\dfrac{1}{2-\sqrt{3}}$的整数部分为$x$，小数部分为$y$，求$\dfrac{x}{y}$的值.
$\dfrac{3}{\sqrt{5}-\sqrt{2}}=\dfrac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}=\dfrac{3(\sqrt{5}+\sqrt{2})}{3}=\sqrt{5}+\sqrt{2}$.
(1)用上述类似的方法化简下列各式：
①$\dfrac{1}{\sqrt{3}+\sqrt{2}}=$
$\sqrt{3}-\sqrt{2}$
；②计算：$\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dots+\dfrac{1}{\sqrt{99}+\sqrt{100}}$；
③计算：$\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dots+\dfrac{1}{\sqrt{99}-\sqrt{100}}$.
(2)已知实数$\dfrac{1}{2-\sqrt{3}}$的整数部分为$x$，小数部分为$y$，求$\dfrac{x}{y}$的值.
答案
11.(1)①$\sqrt{3}-\sqrt{2}$;②$9$;③$-11$ (2)$\dfrac{3\sqrt{3}+3}{2}$
