a) Đặt \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(\Rightarrow A^2=7+4\sqrt{3}+2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+7-4\sqrt{3}\)
\(=14+2\sqrt{49-48}\)
\(=14+2\)
\(=16\)
\(\Rightarrow\left|A\right|=4\)
Mà A > 0 nên suy ra A = 4.
\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{2^2-2.2\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)