题目观察下列一组等式,然后解答问题:(2+1)(2−1)=1(\sqrt{2}+1)(\sqrt{2}-1)=1(2+1)(2−1)=1,(3+2)(3−2)=1(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1(3+2)(3−2)=1,(4+3)(4−3)=1(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})=1(4+3)(4−3)=1,(5+4)(5−4)=1(\sqrt{5}+\sqrt{4})(\sqrt{5}-\sqrt{4})=1(5+4)(5−4)=1,……\ldots \ldots……(1)(1)(1)利用上面的规律,计算:12+1+13+2+14+3+⋯+12024+2023\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+⋯+\frac{1}{\sqrt{2024}+\sqrt{2023}}2+11+3+21+4+31+⋯+2024+20231；(2)(2)(2)请利用上面的规律,比较99−98\sqrt{99}-\sqrt{98}99−98与98−97\sqrt{98}-\sqrt{97}98−97的大小.知识点：最简二次根式章节：数与式 / 二次根式
