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Question-146606




Question Number 146606 by iloveisrael last updated on 14/Jul/21
Answered by som(math1967) last updated on 14/Jul/21
x+(√(x^2 −1))+(1/(x−(√(x^2 −1))))=20  ((x^2 −x^2 +1+1)/(x−(√(x^2 −1))))=20  (1/(x−(√(x^2 −1))))=10  x−(√(x^2 −1))=(1/(10))  again (1/(x−(√(x^2 −1))))=10  ((x+(√(x^2 −1)))/(x^2 −x^2 +1))=10  x+(√(x^2 −1))=10  ∴2x=10+(1/(10))=((101)/(10))  x=((101)/(20))  now x^2 +(√(x^4 −1))+(1/(x^2 +(√(x^4 −1))))  x^2 +(√(x^4 −1)) +((x^2 −(√(x^4 −1)))/(x^4 −x^4 +1))  =2x^2 =2×(((101)/(20)))^2
$${x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\ $$$$\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\ $$$$\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{10} \\ $$$${x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${again}\:\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{10} \\ $$$$\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{10} \\ $$$${x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{10} \\ $$$$\therefore\mathrm{2}{x}=\mathrm{10}+\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{101}}{\mathrm{10}} \\ $$$${x}=\frac{\mathrm{101}}{\mathrm{20}} \\ $$$${now}\:{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}} \\ $$$${x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}\:+\frac{{x}^{\mathrm{2}} −\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}^{\mathrm{4}} −{x}^{\mathrm{4}} +\mathrm{1}} \\ $$$$=\mathrm{2}{x}^{\mathrm{2}} =\mathrm{2}×\left(\frac{\mathrm{101}}{\mathrm{20}}\right)^{\mathrm{2}} \\ $$

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