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Solve-for-real-numbers-x-2-3n-2-4n-2-4-2-n-y-1-x-2-3n-2-4n-2-4-2-n-z-1-x-2-3n-2-4n-2-4-2-n-x-1-n-0-fixed-




Question Number 160712 by HongKing last updated on 05/Dec/21
Solve for real numbers:  (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) y-1  ;  (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) z-1                        (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) x-1  n ∈ (0 ; ∞)  fixed
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{y}-\mathrm{1}\:\:;\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{z}-\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{x}-\mathrm{1} \\ $$$$\mathrm{n}\:\in\:\left(\mathrm{0}\:;\:\infty\right)\:\:\boldsymbol{\mathrm{fixed}} \\ $$

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