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Question Number 128518 by shaker last updated on 08/Jan/21

Commented by bramlexs22 last updated on 08/Jan/21

$$3^{\frac{1}{\mathrm{n}}}=\mathrm{u}\Rightarrow 3^{\frac{1}{\mathrm{n}}}+3^{-\frac{1}{\mathrm{n}}}=\frac{{\mathrm{u}}^2+1}{\mathrm{u}}$$$$\mathrm{x}=\frac{{\mathrm{u}}^2+1}{2\mathrm{u}}\Rightarrow \mathrm{x}+\sqrt{1+{\mathrm{x}}^2}=\frac{{\mathrm{u}}^2+1}{2\mathrm{u}}+\sqrt{1+\frac{{({\mathrm{u}}^2+1)}^2}{{4\mathrm{u}}^2}}$$$$=\frac{({\mathrm{u}}^2+1)+\sqrt{{4\mathrm{u}}^2+{({\mathrm{u}}^2+1)}^2}}{2\mathrm{u}}$$$$\mathrm{A}={\left(\frac{(3^{\frac{2}{\mathrm{n}}}+1)+\sqrt{4.3^{\frac{2}{\mathrm{n}}}+{(3^{\frac{2}{\mathrm{n}}}+1)}^2}}{2.3^{\frac{1}{\mathrm{n}}}}\right)}^{\mathrm{n}}$$

Commented by bramlexs22 last updated on 08/Jan/21

$$\mathrm{may}\mathrm{be}\mathrm{x}=\frac{1}{2}(3^{\frac{1}{\mathrm{n}}}-3^{-\frac{1}{\mathrm{n}}})$$

Answered by MJS_new last updated on 08/Jan/21

$$x=\frac{1}{2}(3^{1/n}+3^{-1/n})=\cosh \left(\frac{\ln 3}{n}\right)$$$$B={(x+\sqrt{x^2-1})}^n={(\cosh \frac{\ln 3}{n}+\sinh \frac{\ln 3}{n})}^n=$$$$={\left(3^{1/n}\right)}^n=3$$$$\mathrm{or}$$$$y=\frac{1}{2}(3^{1/n}-3^{-1/n})=\sinh \left(\frac{\ln 3}{n}\right)$$$$A={(x+\sqrt{x^2+1})}^n={(\sinh \frac{\ln 3}{n}+\cosh \frac{\ln 3}{n})}^n=$$$$={\left(3^{1/n}\right)}^n=3$$

Commented by bramlexs22 last updated on 08/Jan/21

$$\mathrm{waw}...\mathrm{amazing}$$

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