搜题，就来5TiKU.com题库【问题】 阅读下面内容，并将问题解决过程补充完整：$\frac{1}{{\sqrt{2}+1}}=\frac{{1×(\sqrt{2}-1)}}{{(\sqrt{2}+1)(\sqrt{2}-1)}}=\sqrt{2}-1$；$\frac{1}{{\sqrt{3}+\sqrt{2}}}=\frac{{1×(\sqrt{3}-\sqrt{2})}}{{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}=\sqrt{3}-\sqrt{2}$；$\ldots $$\frac{1}{{\sqr
阅读下面内容，并将问题解决过程补充完整：$\frac{1}{{\sqrt{2}+1}}=\frac{{1×(\sqrt{2}-1)}}{{(\sqrt{2}+1)(\sqrt{2}-1)}}=\sqrt{2}-1$；$\frac{1}{{\sqrt{3}+\sqrt{2}}}=\frac{{1×(\sqrt{3}-\sqrt{2})}}{{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}=\sqrt{3}-\sqrt{2}$；$\ldots $$\frac{1}{{\sqrt{100}+\sqrt{99}}}=\frac{{1×(\sqrt{100}-\sqrt{99})}}{{(\sqrt{100}+\sqrt{99})(\sqrt{100}-\sqrt{99})}}=\sqrt{100}-\sqrt{99}$.由此，我们可以解决下面问题：$S=1+\frac{1}{{\sqrt{2}}}+\frac{1}{{\sqrt{3}}}+\ldots +\frac{1}{{\sqrt{100}}}$，请求出$S$的整数部分.解：$S=1+\frac{1}{{\sqrt{2}}}+\frac{1}{{\sqrt{3}}}+\ldots +\frac{1}{{\sqrt{100}}}=\frac{2}{{1+1}}+\frac{2}{{\sqrt{2}+\sqrt{2}}}+\frac{2}{{\sqrt{3}+\sqrt{3}}}+\ldots +\frac{1}{{\sqrt{100}+\sqrt{100}}}$$ \lt \frac{2}{{1+1}}+\frac{2}{{1+\sqrt{2}}}+\frac{2}{{\sqrt{2}+\sqrt{3}}}+\ldots +\frac{2}{{\sqrt{99}+\sqrt{100}}}=1+2(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots +\sqrt{100}-\sqrt{99})=19$；$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots +\frac{1}{\sqrt{100}}=\frac{2}{1+1}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+\ldots +\frac{2}{\sqrt{100}+\sqrt{100}} \gt \_\_\_\_\_\_.$$\therefore S$的整数部分是______.
正确答案：$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{100}}=\frac{2}{1+1}+\frac{2}{\sqrt{2}+\sqrt{2}}+\cdots +\frac{2}{\sqrt{100}+\sqrt{100}} \lt \frac{2}{1+1}+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\cdots +\frac{2}{\sqrt{99}+\sqrt{1
题目解析：
根据题中给出的计算方法进行计算求解即可.
点评：
此题考查了估算无理数的大小，正确估算出$S$的整数部分是$18$是解题的关键.
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