阅读下面问题： $\dfrac{1}{ \sqrt{2}+1} = \dfrac{1×\left( \sqrt{2}-1\right)}{\left( \sqrt{2}+1\right)\left( \sqrt{2}-1\right)} = \sqrt{2}-1$； $\dfrac{1}{ \sqrt{3}+ \sqrt{2}} = \dfrac{ \sqrt{3}- \sqrt{2}}{\left( \sqrt{3}+ \sqrt{2}\right)\left( \sqrt{3}- \sqrt{2}\right)} = \sqrt{3}- \sqrt{2} ; \dfrac{1}{ \sqrt{5}+2} = \dfrac{ \sqrt{5}-2}{\left( \sqrt{5}+2\right)\left( \sqrt{5}+2\right)} = \sqrt{5} -2$；$…$ 试求： $(1) \dfrac{1}{3+2 \sqrt{2}}$的值； $(2) \dfrac{1}{ \sqrt{n+1}+ \sqrt{n}} (n$为正整数$)$的值； $(3) \dfrac{1}{ \sqrt{2}+1} + \dfrac{1}{ \sqrt{3}+ \sqrt{2}} + \dfrac{1}{ \sqrt{4}+ \sqrt{3}} +⋯+ \dfrac{1}{ \sqrt{100}+ \sqrt{99}}$的值．

1. 将分式分子分母同乘以分母的共轭因式以进行分母有理化

$\dfrac{1}{3+2 \sqrt{2}} = \dfrac{3-2 \sqrt{2}}{\left(3+2 \sqrt{2}\right)\left(3-2 \sqrt{2}\right)}$

2. 利用平方差公式简化分母并得出最终结果

$= \dfrac{3-2 \sqrt{2}}{3^2 - \left(2 \sqrt{2}\right)^2} = \dfrac{3-2 \sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}$

1. 将含字母 $n$ 的分式进行分母有理化

$\dfrac{1}{\sqrt{n+1} + \sqrt{n}} = \dfrac{\sqrt{n+1} - \sqrt{n}}{\left(\sqrt{n+1} + \sqrt{n}\right)\left(\sqrt{n+1} - \sqrt{n}\right)}$

2. 利用平方差公式化简含有字母的分母

$= \dfrac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} = \dfrac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$

1. 应用第(2)问的通项有理化结论，将求和式中的每一项展开

根据第(2)问的结论 $\dfrac{1}{\sqrt{n+1}+\sqrt{n}} = \sqrt{n+1}-\sqrt{n}$，原式可变形为：
$\left(\sqrt{2}-1\right) + \left(\sqrt{3}-\sqrt{2}\right) + \left(\sqrt{4}-\sqrt{3}\right) + \dots + \left(\sqrt{100}-\sqrt{99}\right)$

2. 通过抵消中间项进行裂项相消计算并得出最终结果

观察式子，中间的项可以两两抵消：
$= -1 + \left(\sqrt{2}-\sqrt{2}\right) + \left(\sqrt{3}-\sqrt{3}\right) + \dots + \left(\sqrt{99}-\sqrt{99}\right) + \sqrt{100}$
$= -1 + \sqrt{100}$
$= -1 + 10 = 9$