Nhận thấy:  \(y>2\)

Ta xét:

\(y^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}\)

\(\Rightarrow\)  \(\left(y^2-5\right)^2=13+y\)

\(\Leftrightarrow\)  \(y^4-10y^2-y+12=0\)

\(\Leftrightarrow\)  \(\left(y^4-9y^2\right)-\left(y^2-9\right)-\left(y-3\right)=0\)

\(\Leftrightarrow\)  \(\left[\left(y+3\right)\left(y+1\right)\left(y-1\right)-1\right]\left(y-3\right)=0\)

Mà  \(y>2\)  nên  \(\left[\left(y+3\right)\left(y+1\right)\left(y-1\right)-1\right]>0\)

Do đó, ta dễ dàng suy ra  \(y-3=0\)  hay  \(y=3\)