The main idea of the exercise is to use the relation $a^{2}-b^{2}=(a-b)(a+b)$ to write: $1=(\sqrt{k+1}+\sqrt{k}) \cdot(\sqrt{k+1}-\sqrt{k})$, or equivalently $\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k}$.

In other words, $\frac{1}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}, \frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2} \cdots \frac{1}{\sqrt{99}+\sqrt{100}}=\sqrt{100}-\sqrt{99}$. This leads to:

$$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}=(-\sqrt{1}+\sqrt{2})+(-\sqrt{2}+\sqrt{3})+\cdots$$

$\cdots+(-\sqrt{k}+\sqrt{k+1})+(-\sqrt{99}+\sqrt{100})=-\sqrt{1}+\sqrt{100}=9$ because all the other square roots cancel out ("telescoping simplifications").

Whenever we see square roots in a sum in the denominator, we should have the reflex to eliminate them using this same technique, by multiplying the numerator and the denominator by what is called the "conjugate expression."