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Question Number 210291 by Ismoiljon_008 last updated on 05/Aug/24

Answered by mr W last updated on 06/Aug/24

$$x=\sqrt{x^2}=\sqrt{2^2+x^2-2^2}=\sqrt{2^2+(x-2)(x+2)}$$$$=\sqrt{2^2+(x-2)\sqrt{2^2+x(x+4)}}$$$$=\sqrt{2^2+(x-2)\sqrt{2^2+x\sqrt{2^2+(x+2)(x+6)}}}$$$$=\sqrt{2^2+(x-2)\sqrt{2^2+x\sqrt{2^2+(x+2)\sqrt{2^2+(x+4)(x+8)}}}}$$$$=\sqrt{2^2+(x-2)\sqrt{2^2+x\sqrt{2^2+(x+2)\sqrt{2^2+(x+4)\sqrt{2^2+...}}}}}$$$$withx=5:$$$$5=\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{4+9\sqrt{4+...}}}}}$$

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