27. (10 分)阅读下面解题过程,然后回答问题.
$\frac{1}{\sqrt{5}+\sqrt{4}}= \frac{1 ×(\sqrt{5}-\sqrt{4})}{(\sqrt{5}+\sqrt{4}) ×(\sqrt{5}-\sqrt{4})}= \frac{(\sqrt{5}-\sqrt{4})}{(\sqrt{5})^{2}-(\sqrt{4})^{2}}= \sqrt{5}-\sqrt{4}= \sqrt{5}-2$；
$\frac{1}{\sqrt{6}+\sqrt{5}}= \frac{1 ×(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5}) ×(\sqrt{6}-\sqrt{5})}= \frac{\sqrt{6}-\sqrt{5}}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}= \sqrt{6}-\sqrt{5}$.
(1)观察上面的解题过程,请直接写出式子：$\frac{1}{\sqrt{n}+\sqrt{n-1}}=$.
(2)利用上面所提供的解法,请化简：
$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+…+\frac{1}{\sqrt{10}+\sqrt{9}}$.

(1)
$\frac{1}{\sqrt{n}+\sqrt{n - 1}}=\frac{\sqrt{n}-\sqrt{n - 1}}{(\sqrt{n}+\sqrt{n - 1})(\sqrt{n}-\sqrt{n - 1})}=\frac{\sqrt{n}-\sqrt{n - 1}}{n-(n - 1)}=\sqrt{n}-\sqrt{n - 1}$
(2)
$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{10}+\sqrt{9}}$
$=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+(\sqrt{5}-\sqrt{4})+\cdots+(\sqrt{10}-\sqrt{9})$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\cdots+\sqrt{10}-\sqrt{9}$
$=\sqrt{10}-1$