Question

tính giá trị

A= \(\sqrt{21+8\sqrt{ }5}\)+\(\sqrt{21-8\sqrt{ }5}\)

Answer 1

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 \(4-\sqrt{5}\)

Answer 2

\(A=\sqrt{21+8\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)

      \(=8\)

Answer 3

\(A=\sqrt{21+8\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}\right)^2+8\sqrt{5}+4^4}+\sqrt{\left(\sqrt{5}\right)^2-8\sqrt{5}+4^2}\)

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

\(A=\left|\sqrt{5}+4\right|+\left|\sqrt{5}-4\right|\)

\(A=\sqrt{5}+4+4-\sqrt{5}\) (   vi  \(\sqrt{5}-4< 0\)  )

\(A=8\)

vay \(A=8\)

Answer 4

\(A^2=21+8\sqrt{5}+21-8\sqrt{5}+2\sqrt{21^2-8^2.5}\)

\(=42+2\sqrt{441-320}\)

\(=42+2\sqrt{121}\)

\(=42+2.11\)

\(=64\)

Mà \(A>0\)

\(\Rightarrow A=8\)