tui làm 1 cái thui nha
\(\sqrt{21+8\sqrt{5}}\)
=\(\sqrt{16+2\cdot4\cdot\sqrt{5}+5}\)
=\(\sqrt{\left(4+\sqrt{5}\right)^2}\)
=\(4+\sqrt{5}\)
cái kia \(4-\sqrt{5}\)
\(A=\sqrt{21+8\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(=8\)
\(A=\sqrt{21+8\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}\right)^2+8\sqrt{5}+4^4}+\sqrt{\left(\sqrt{5}\right)^2-8\sqrt{5}+4^2}\)
\(A=\sqrt{\left(\sqrt{5}+4\right)^2}+\sqrt{\left(\sqrt{5}-4\right)^2}\)
\(A=\left|\sqrt{5}+4\right|+\left|\sqrt{5}-4\right|\)
\(A=\sqrt{5}+4+4-\sqrt{5}\) ( vi \(\sqrt{5}-4< 0\) )
\(A=8\)
vay \(A=8\)
\(A^2=21+8\sqrt{5}+21-8\sqrt{5}+2\sqrt{21^2-8^2.5}\)
\(=42+2\sqrt{441-320}\)
\(=42+2\sqrt{121}\)
\(=42+2.11\)
\(=64\)
Mà \(A>0\)
\(\Rightarrow A=8\)
