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Question-210291




Question Number 210291 by Ismoiljon_008 last updated on 05/Aug/24
Answered by mr W last updated on 06/Aug/24
x=(√x^2 )=(√(2^2 +x^2 −2^2 ))=(√(2^2 +(x−2)(x+2)))  =(√(2^2 +(x−2)(√(2^2 +x(x+4)))))  =(√(2^2 +(x−2)(√(2^2 +x(√(2^2 +(x+2)(x+6)))))))  =(√(2^2 +(x−2)(√(2^2 +x(√(2^2 +(x+2)(√(2^2 +(x+4)(x+8)))))))))  =(√(2^2 +(x−2)(√(2^2 +x(√(2^2 +(x+2)(√(2^2 +(x+4)(√(2^2 +...))))))))))  with x=5:  5=(√(4+3(√(4+5(√(4+7(√(4+9(√(4+...))))))))))
$${x}=\sqrt{{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +{x}\left({x}+\mathrm{4}\right)}} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}+\mathrm{2}\right)\left({x}+\mathrm{6}\right)}}} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}+\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}+\mathrm{4}\right)\left({x}+\mathrm{8}\right)}}}} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}+\mathrm{2}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}+\mathrm{4}\right)\sqrt{\mathrm{2}^{\mathrm{2}} +…}}}}} \\ $$$${with}\:{x}=\mathrm{5}: \\ $$$$\mathrm{5}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{4}+\mathrm{5}\sqrt{\mathrm{4}+\mathrm{7}\sqrt{\mathrm{4}+\mathrm{9}\sqrt{\mathrm{4}+…}}}}} \\ $$

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