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Question Number 212006 by RojaTaniya last updated on 26/Sep/24

	Answered by a.lgnaoui last updated on 26/Sep/24
	$posons z=x+100 (((z−2)^5 +(z+2)^5 )/((z−1)^5 +(z+1)^5 ))=((2z^5 +80z^3 +160z)/(2z^5 +20z^3 +10z)) =((z^4 +40z^2 +80)/(z^4 +10z^2 +5))=((16^2 )/(11^2 −60))=((256)/(61)) alors 61z^4 +2440z^2 +4880=256z^4 +2560z^2 +128 256z^4 2560z^2 +1280 13z^4 +8z^2 −240=0 16+3120=56^2 z^2 =4 z={−2,+2} { ((x=−98)),((x=−102)) :}$
	$posons z = x + 100$ $\frac{{(z - 2)}^{5} + {(z + 2)}^{5}}{{(z - 1)}^{5} + {(z + 1)}^{5}} = \frac{2 z^{5} + 80 z^{3} + 160 z}{2 z^{5} + 20 z^{3} + 10 z}$ $= \frac{z^{4} + 40 z^{2} + 80}{z^{4} + 10 z^{2} + 5} = \frac{16^{2}}{11^{2} - 60} = \frac{256}{61}$ $alors$ $61 z^{4} + 2440 z^{2} + 4880 = 256 z^{4} + 2560 z^{2} + 128$ $256 z^{4} 2560 z^{2} + 1280$ ${13 z}^{4} + {8 z}^{2} - 240 = 0$ $16 + 3120 = 56^{2}$ $z^{2} = 4 z = {- 2, + 2}$ ${\begin{cases} x = - 98 \\ x = - 102 \end{cases}$
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