đặt A=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(A^2=8+2\sqrt{16-10-2\sqrt{5}}\)
\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(A^2=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
vậy A=\(\sqrt{5}+1\)