Giải thích các bước giải:
Ta có:
$\begin{split}\dfrac{1}{\sqrt{k-1}+\sqrt{k}}&=\dfrac{k-(k-1)}{\sqrt{k-1}+\sqrt{k}}\\&=\dfrac{(\sqrt{k}+\sqrt{k-1})(\sqrt{k}-\sqrt{k-1})}{\sqrt{k-1}+\sqrt{k}}\\&=\sqrt{k}-\sqrt{k-1}\end{split}$
$\rightarrow \begin{cases}\dfrac{1}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}\\\dfrac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\\...\\\dfrac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n}-\sqrt{n-1}\end{cases}$
$\rightarrow\dfrac{1}{\sqrt{1}+\sqrt{2}}+ \dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n}-1$
