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Question Number 148229 by iloveisrael last updated on 26/Jul/21

Commented by nimnim last updated on 26/Jul/21

$$x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20$$$$\Rightarrow \frac{x^2-(x^2-1)+1}{x-\sqrt{x^2-1}}\times \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=20$$$$\Rightarrow \frac{2(x+\sqrt{x^2-1})}{x^2-(x^2-1)}=20\iff x+\sqrt{x^2-1}=10$$$$\Rightarrow x^2-1=100-20x+x^2\iff x=\frac{101}{20}$$$$Now,x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}$$$$=x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}\times \frac{x^2-\sqrt{x^4-1}}{x^2-\sqrt{x^4-1}}$$$$=x^2+\sqrt{x^4-1}+\frac{x^2-\sqrt{x^4-1}}{x^4-(x^4-1)}$$$$=x^2+\sqrt{x^4-1}+x^2-\sqrt{x^4-1}$$$$=2x^2=2{\left(\frac{101}{20}\right)}^2=\frac{10201}{200}=51.005★$$$$$$

Answered by EDWIN88 last updated on 26/Jul/21

$$\left(1\right)(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})+1=20(x-\sqrt{x^2-1})$$$$\Rightarrow 2=20(x-\sqrt{x^2-1})$$$$\Rightarrow \sqrt{x^2-1}=x-\frac{1}{10}$$$$\Rightarrow x^2-1=x^2-\frac{x}{5}+\frac{1}{100}$$$$\Rightarrow \frac{x}{5}=\frac{101}{100},x=\frac{101}{20}$$$$\left(2\right)x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=?$$$$(\bullet )x^2-1+\sqrt{(x^2-1)(x^2+1)}+1$$$$=(x+1)(x-1)+\sqrt{(x+1)(x-1)(x^2+1)}+1$$$$=\left(\frac{121}{20}\right)\left(\frac{81}{20}\right)+\sqrt{\left(\frac{121}{20}\right)\left(\frac{81}{20}\right)\left(\frac{{101}^2+{20}^2}{{20}^2}\right)}+1$$$$=\frac{9801}{400}+\frac{99\sqrt{10601}}{400}+1$$$$=\frac{99\sqrt{10601}+10201}{400}$$$$so{x}^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=$$$$\frac{99\sqrt{10601}+10201}{400}+\frac{400}{99\sqrt{10601}+10201}$$$$\left[loveJew\right]$$

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