\(A=\frac{10}{3}+\frac{10}{9}+\frac{10\sqrt{5}}{3\sqrt{36\sqrt{6}}}\)

\(A=\frac{40}{9}+\frac{10\sqrt{5}}{18\sqrt{\sqrt{6}}}\)

trục căn thứa là ra nha bạn

\(A=\frac{40}{9}+\frac{10\sqrt{5}}{18\sqrt{\sqrt{6}}}\)

B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

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

D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)

với n >0, ta có :

\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)

Gọi biểu thức đã cho là A

\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)

\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)

\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)

\(A=-\sqrt{1}+\sqrt{9}=2\)

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

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

Ta có:

\(A=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{5}{12}-\frac{1}{16}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{17}{48}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{51}}{12}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{17}}{12}\)

\(A=\frac{4\sqrt{3}+2\sqrt{2}+\sqrt{17}}{12}\)

Ta có: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}=\sqrt{\frac{5-2\sqrt{6}}{12}}\)

Vì \(5-2\sqrt{6}=3-2\sqrt{3}.\sqrt{2}+2=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\)\(\Rightarrow5-2\sqrt{6}=\left(\sqrt{3}-\sqrt{2}\right)^2\)

Như vậy: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)

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

Rút gọn ta được \(A=\frac{\sqrt{3}}{2}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5-2\sqrt{6}}{12}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}\)

\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}}\left(do\sqrt{3}-\sqrt{2}>0\right)\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{6}\left(\sqrt{3}-\sqrt{2}\right)=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}-\frac{\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

Phân tích mỗi hạng tử theo kiểu như dưới đây

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

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

Khi đó mọi mẫu đều bằng -1

Bạn tiếp tục làm và kết quả nhận được là \(1-\sqrt{9}\)