Question

\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

Answer 1

Lời giải:

Đặt \(a=\sqrt{4+\sqrt{10+2\sqrt{5}}}; b\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(a^2+b^2=8; ab=\sqrt{4^2-(10+2\sqrt{5})}=\sqrt{6-2\sqrt{5}}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)

\(A=a+b\)

\(A^2=a^2+b^2+2ab=8+2(\sqrt{5}-1)=6+2\sqrt{5}=(\sqrt{5}+1)^2\)

Vì $A>0$ nên $A=\sqrt{5}+1$