\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)

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

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)

\(=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}\\ =\dfrac{\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)}{\sqrt{2}}\\ =\dfrac{\left(\sqrt{7+2\sqrt{7}.\sqrt{3}+3}-\sqrt{7-2\sqrt{7}.\sqrt{3}+3}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{3}{2}}=2\sqrt{\dfrac{3}{2}}=\sqrt{6}\)

Lời giải:
Đặt biểu thức là $A$. Ta có:

\(A=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{2}.\sqrt{5-\sqrt{21}}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{10-2\sqrt{21}}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{(\sqrt{7}-\sqrt{3})^2}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})|\sqrt{7}-\sqrt{3}|=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})^2\)

\(=(5+\sqrt{21})(10-2\sqrt{21})=2(5+\sqrt{21})(5-\sqrt{21})=2(5^2-21)=8\)

Ta có: \(\left(5+\sqrt{21}\right)\cdot\left(\sqrt{14}-\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)

\(=\dfrac{\left(10+2\sqrt{21}\right)\cdot\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}}{2}\)

\(=\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2\cdot\left(\sqrt{7}-\sqrt{3}\right)^2}{2}\)

=8

1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)

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

\(=2\sqrt{2}-1\)

2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)

\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)

\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)

\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)

\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)

\(=2\sqrt{5}-1+2\sqrt{5}+1\)

\(=4\sqrt{5}\)

Tham khảo ạ

\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)

\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)

\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)

\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)

                      \(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)

                     \(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)

\(\Rightarrow\)\(B=\sqrt{22}\)

cho mk hỏi căn viết thế nào

\(\sqrt{\sqrt{5}-\sqrt{5-\sqrt{21-4\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\sqrt{20^2}-2.\sqrt{20}+1}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\left(\sqrt{20}-1\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\left|\sqrt{20}-1\right|}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{20}+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{5^2}-2\sqrt{5}+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

\(=1\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

\(\sqrt{21-8\sqrt{5}}\)\(-\sqrt{21-4\sqrt{5}}\)

\(=\sqrt{16-2.4\sqrt{5}+5}\)\(-\sqrt{20-2\sqrt{20}+1}\)

\(=\sqrt{\left(4-\sqrt{5}\right)^2}\)\(-\sqrt{\left(\sqrt{20}-1\right)}\)

\(=4-\sqrt{5}-\left(\sqrt{20}-1\right)\)

\(=4-\sqrt{5}-\sqrt{20}+1\)

\(=5-\sqrt{5}-2\sqrt{5}\)

\(=5-3\sqrt{5}\)

\(A=\)bn ghi lại đề nha mình lười

\(=\left(\sqrt{5+\sqrt{21}}\right)^2\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{5-\sqrt{21}}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)

\(=\left(\sqrt{\left(5^2-21\right)}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)

\(=2.\left(\sqrt{5+\sqrt{21}}\right)\sqrt{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\left(\sqrt{10+2\sqrt{21}}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\left(\sqrt{7+2\sqrt{21}+3}\right) \left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=2.\left(7-3\right)=2.4=8\)

tíck mình nha bn thanks nhìu !!!!!!!!!