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\displaystyle [br]\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}
\displaystyle [br]\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}
\displaystyle[br]\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}
\displaystyle [br]\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}
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\displaystyle [br]\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}
\displaystyle [br]\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}
\displaystyle[br]\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}
\displaystyle [br]\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}
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