题目观察下列各式，通过分母有理化把不是最简二次根式的化成最简二次根式．12+1=1×(2−1)(2+1)(2−1)=2−1(2)2−1=2−12−1=2−1;\dfrac{1}{\sqrt{2}+1}=\dfrac{1\text{×}(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^{2}-1}=\dfrac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\text{;}2+11=(2+1)(2−1)1×(2−1)=(2)2−12−1=2−12−1=2−1;13+2=1×(3−2)(3+2)(3−2)=3−2(3)2−(2)2=3−23−2=3−2.\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{1\text{×}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}.3+21=(3+2)(3−2)1×(3−2)=(3)2−(2)23−2=3−23−2=3−2.按照以上的过程，解答以下问题：(1)(1)(1)计算：14+3;\dfrac{1}{\sqrt{4}+\sqrt{3}}\text{;}4+31;(2)(2)(2)计算：(12+1+13+2+14+3+⋯+12021+2020)×(2021+1).(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\text{+⋯+}\dfrac{1}{\sqrt{2021}+\sqrt{2020}})\text{×}(\sqrt{2021}+1).(2+11+3+21+4+31+⋯+2021+20201)×(2021+1).知识点：平方差公式、二次根式分母有理化、二次根式的混合运算章节：第21章 二次根式 / 21.3 二次根式的加减
