16 观察下列运算:
①由$(\sqrt{2}+1) × (\sqrt{2}-1)=1$,得$\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$;
②由$(\sqrt{3}+\sqrt{2}) × (\sqrt{3}-\sqrt{2})=1$,得$\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}$;
③由$(\sqrt{4}+\sqrt{3}) × (\sqrt{4}-\sqrt{3})=1$,得$\frac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}$;
……
(1)通过观察你得出什么规律? 用含$n$的式子表示出来.
(2)利用(1)中你发现的规律计算:
$(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+···+\frac{1}{\sqrt{2024}+\sqrt{2023}}) × (\sqrt{2024}+1)$

解:(1)由题意,得$\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$($n$为正整数).
(2)$(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+…+\dfrac{1}{\sqrt{2024}+\sqrt{2023}})×(\sqrt{2024}+1)$
$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+…+\sqrt{2024}-\sqrt{2023})×(\sqrt{2024}+1)$
$=(\sqrt{2024}-1)×(\sqrt{2024}+1)$
$=(\sqrt{2024})^2-1$
$=2024-1$
$=2023$.