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Question-49359




Question Number 49359 by peter frank last updated on 06/Dec/18
Answered by MJS last updated on 06/Dec/18
it′s wrong  y=(1/x) ⇒ (((x^2 −x+1)^2 )/(x(x−1)^2 ))=(((y^2 −y+1)^2 )/(y(y−1)^2 )); (y^2 /(y−1))=(1/(x(1−x)))  anyway the equation is  (x^2 −3x+1)^2 =0 ⇒ x=(3/2)±((√5)/2)
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$$${y}=\frac{\mathrm{1}}{{x}}\:\Rightarrow\:\frac{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\left({y}^{\mathrm{2}} −{y}+\mathrm{1}\right)^{\mathrm{2}} }{{y}\left({y}−\mathrm{1}\right)^{\mathrm{2}} };\:\frac{{y}^{\mathrm{2}} }{{y}−\mathrm{1}}=\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$\mathrm{anyway}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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