Question

6. Cho \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

a) Tính \(A^2\)

b) Rút gọn A

Answer 1

a) \(A^2=\left(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\right)^2\)

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

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

\(=4^2\)

\(=16\)

b) \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

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

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

\(=4\)

Answer 2

a, \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\Rightarrow7+4\sqrt{3}+7-4\sqrt{3}=14\)