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Question Number 147252 by mathlove last updated on 19/Jul/21

Answered by Olaf_Thorendsen last updated on 19/Jul/21

$$l={(\frac{1}{5}+{(\frac{1}{5}+{(\frac{1}{5}+...)}^2)}^2)}^2$$$$l={(\frac{1}{5}+l)}^2$$$$l^2-\frac{3}{5}l+\frac{1}{25}=0$$$$l=\frac{\frac{3}{5}\pm \sqrt{\frac{9}{25}-4\times 1\times \frac{1}{25}}}{2}$$$$l=\frac{\frac{3}{5}\pm \frac{1}{\sqrt{5}}}{2}=\frac{1}{10}(3\pm \sqrt{5})$$$$l>{\left(\frac{1}{5}\right)}^2=0,25\Rightarrow \frac{1}{10}(3-\sqrt{5})\mathrm{impossible}$$$$l=\frac{1}{10}(3+\sqrt{5})$$$$\mathrm{X}=3-10l=-\sqrt{5}$$

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