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2. 已知一个长方形的长和宽之比为5:3，它的周长为 $ 1 6 \sqrt{5} \mathrm{c m} $ ，求这个长方形的面积.
答案
2. 依题意可设这个长方形的长和宽分别为$5x$,$3x$,则$2(5x+3x)=16\sqrt{5}$,解得$x=\sqrt{5}$. 所以,长、宽分别为$5\sqrt{5}\mathrm{cm}$,$3\sqrt{5}\mathrm{cm}$. 则它的面积为$5\sqrt{5} × 3\sqrt{5}=75(\mathrm{cm^{2}})$.
3. * 阅读下列材料：
$\begin{array}{l} \frac {1}{\sqrt {2} + 1} = \frac {1 × (\sqrt {2} - 1)}{(\sqrt {2} + 1) (\sqrt {2} - 1)} = \sqrt {2} - 1; \frac {1}{\sqrt {3} + \sqrt {2}} = \frac {1 × (\sqrt {3} - \sqrt {2})}{(\sqrt {3} + \sqrt {2}) (\sqrt {3} - \sqrt {2})} = \sqrt {3} - \sqrt {2}; \\ \frac {1}{\sqrt {5} + 2} = \frac {\sqrt {5} - 2}{(\sqrt {5} + 2) (\sqrt {5} - 2)} = \sqrt {5} - 2. \\ \end{array}$
请根据以上的计算方法求：
$\begin{array}{l} \frac {1}{\sqrt {2} + 1} = \frac {1 × (\sqrt {2} - 1)}{(\sqrt {2} + 1) (\sqrt {2} - 1)} = \sqrt {2} - 1; \frac {1}{\sqrt {3} + \sqrt {2}} = \frac {1 × (\sqrt {3} - \sqrt {2})}{(\sqrt {3} + \sqrt {2}) (\sqrt {3} - \sqrt {2})} = \sqrt {3} - \sqrt {2}; \\ \frac {1}{\sqrt {5} + 2} = \frac {\sqrt {5} - 2}{(\sqrt {5} + 2) (\sqrt {5} - 2)} = \sqrt {5} - 2. \\ \end{array}$
请根据以上的计算方法求：
答案
3. (1)$\sqrt{7}+\sqrt{6}$ (2)$2\sqrt{3}-\sqrt{11}$
