阅读下面材料,并解决问题:
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$；
$\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}$；
$\ldots \ldots$
$(1)$填空:$\frac{1}{\sqrt{11}+\sqrt{10}}=$______；
$(2)$猜想:当$n$是正整数时,$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\_\_\_\_\_\_;($用含$n$的式子表示）
$(3)$计算:$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{2+\sqrt{3}}+\ldots +\frac{1}{\sqrt{2023}+\sqrt{2022}}=\_\_\_\_\_\_.$