Đặt \(A=\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3\)
\(A^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}\)
\(A^2=6+A\)
\(A^2-3A+2A-6=0\)
\(A\left(A-3\right)+2\left(A-3\right)=0\)
\(\Rightarrow\left(A-3\right)\left(A+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}A-3=0\\A+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=3\left(n\right)\\A=-2\left(l\right)\end{cases}}}\)
Vậy \(\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3\left(đpcm\right)\)