Đặt \(a=\sqrt{4-\sqrt{10+2\sqrt{5}}};b=\sqrt{4+\sqrt{10+2\sqrt{5}}}\)(a; b >0)
Xét (a + b)2= a2 + b2 + 2ab
= \(4-\sqrt{10+2\sqrt{5}}+4+\sqrt{10+2\sqrt{5}}\)+
\(2.\sqrt{\left(4-\sqrt{10+2\sqrt{5}}\right)\left(4+\sqrt{10+2\sqrt{5}}\right)}\)
= \(8+2.\sqrt{6-2\sqrt{5}}=8+2.\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
= \(\left(\sqrt{5}+1\right)^2\)=> a + b = \(\sqrt{5}+1\)