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Question Number 190925 by sonukgindia last updated on 14/Apr/23

Commented by Frix last updated on 14/Apr/23

$$x=\frac{1+\sqrt{5}}{2}$$

Answered by ibrahimabdullayev last updated on 14/Apr/23

$$$$$$Solution:IbrahimAbdullayev$$$${(x-\frac{1}{x})}^{\frac{1}{2}}-{(1-\frac{1}{x})}^{\frac{1}{2}}=\frac{x-1}{x}$$$${(x-\frac{1}{x})}^{\frac{1}{2}}+{(1-\frac{1}{x})}^{\frac{1}{2}}=x$$$$2{(x-\frac{1}{x})}^{\frac{1}{2}}=\frac{x^2+x-1}{x}$$$$\frac{4x^2-4}{x}=\frac{{(x^2+x-1)}^2}{x^2}$$$$x^4-2x^3-x^2+2x+1=0$$$$x^2+\frac{1}{x^2}-2(x-\frac{1}{x})-1=0$$$$x-\frac{1}{x}=t\Rightarrow t^2-2t+1=0\Rightarrow t=1$$$$x-\frac{1}{x}=1\Rightarrow x^2-x-1=0\Rightarrow x=\frac{1+\sqrt{5}}{2}$$$$x=\frac{1+\sqrt{5}}{2}$$

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