题目阅读下面材料,并解决问题:12+1=2−1(2+1)(2−1)=2−1\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-12+11=(2+1)(2−1)2−1=2−1；13+2=3−2(3+2)(3−2)=3−2\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}3+21=(3+2)(3−2)3−2=3−2；12+3=2−3(2+3)(2−3)=2−3\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}2+31=(2+3)(2−3)2−3=2−3；……\ldots \ldots……(1)(1)(1)填空:111+10=\frac{1}{\sqrt{11}+\sqrt{10}}=11+101=______；(2)(2)(2)猜想:当nnn是正整数时,1n+1+n=______;(\frac{1}{\sqrt{n+1}+\sqrt{n}}=\_\_\_\_\_\_;(n+1+n1=______;(用含nnn的式子表示）(3)(3)(3)计算:12+1+13+2+12+3+…+12023+2022=______.\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{2+\sqrt{3}}+\ldots +\frac{1}{\sqrt{2023}+\sqrt{2022}}=\_\_\_\_\_\_.2+11+3+21+2+31+…+2023+20221=______.知识点：实数的运算章节：未标注
