Đặt \(x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}>0\)
\(\Leftrightarrow x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}\)
\(\Leftrightarrow x^2=6+x\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x=3\)
Vậy \(\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3\)