Ta có:

\(B=\frac{\sqrt{2+\sqrt{3}}}{2}\div\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}\div\left(\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}\div\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left(\frac{\sqrt{3}+1}{2}-\frac{2\sqrt{3}}{2}+\frac{\left(\sqrt{3}+1\right)\sqrt{3}}{6}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left[\frac{3\left(\sqrt{3}+1\right)-6\sqrt{3}+3+\sqrt{3}}{6}\right]\)

\(B=\frac{\sqrt{3}+1}{2}\div\frac{6-2\sqrt{3}}{6}\)

\(B=\frac{\sqrt{3}+1}{2}.\frac{6}{6-2\sqrt{3}}\)

\(B=\frac{3+2\sqrt{3}}{2}\)

\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)

\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{5\sqrt{6}-12}{18\sqrt{2}}\)

\(E=\frac{36\sqrt{2}}{18\sqrt{6}}+\frac{12\sqrt{3}}{18\sqrt{6}}+\frac{\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{3}}\)

\(E=\frac{36\sqrt{2}+12\sqrt{3}+\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{6}}\)

\(E=\frac{51\sqrt{2}}{18\sqrt{6}}\)

\(E=\frac{17\sqrt{2}}{6\sqrt{6}}\)

\(E=\frac{17\sqrt{2}}{2.3\sqrt{2}.\sqrt{3}}\)

\(E=\frac{17}{\sqrt{2}.3\sqrt{2}.\sqrt{3}}\)

\(E=\frac{17}{6\sqrt{3}}\)

\(E=\frac{17\sqrt{3}}{18}\)

b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào bài toán ta được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{99}-\frac{1}{\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Cả 2 câu là n tự nhiên khác 0 hết nhé

a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

Áp đụng vào bài toán được

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)

\(=\sqrt{1681}-\sqrt{1}=41-1=40\)

a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)

b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)

\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)\(=\frac{4+2\sqrt{3}}{\sqrt{4}+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{\sqrt{4}-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{4+2\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{4-2\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)\(=\frac{4+2\sqrt{3}}{2+\sqrt{3}+1}+\frac{4-2\sqrt{3}}{2-\sqrt{3}+1}\)

\(=\frac{\left(\sqrt{3}+1\right)^2}{3+\sqrt{3}}+\frac{\left(\sqrt{3}-1\right)^2}{3-\sqrt{3}}\)

\(=\frac{\left(\sqrt{3}+1\right)^2}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}\)

\(=\frac{2\sqrt{3}}{\sqrt{3}}=2\)