Đặt \(a=\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}\left(a>0\right)\)
\(\Rightarrow a^2=12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}\)
\(\Rightarrow a^2-a=\)\(12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12}}}}\)\(-\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}-12=0\)
\(\Rightarrow a^2-a-12=0\)
\(\Leftrightarrow\left(a+3\right)\left(a-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-3\left(ktm\right)\\a=4\left(tm\right)\end{matrix}\right.\)
Vậy \(\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}=4\)
Đúng 0
Bình luận (0)
