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Question Number 188107 by pascal889 last updated on 25/Feb/23

	Answered by CElcedricjunior last updated on 25/Feb/23
	$((x/(x−2)))^2 +((x/(x+2)))^2 =2 ∃ssi x≠−2 et x≠2 =>x^2 (x^2 +4x+4)+x^2 (x^2 −4x+4)=2(x^4 −8x^2 +16)★Moivre =>8x^2 =32−16x^2 ★Cedric junior =>24x^2 =32=>x=∓(√(4/3))=∓((2(√3))/3) S_R ={−((2(√3))/3);((2(√3))/3)}$
	${(\frac{x}{x - 2})}^{2} + {(\frac{x}{x + 2})}^{2} = 2$ $\exists s s i x \neq - 2 e t x \neq 2$ $=> x^{2} (x^{2} + 4 x + 4) + x^{2} (x^{2} - 4 x + 4) = 2 (x^{4} - 8 x^{2} + 16) ★ M o i v r e$ $=> 8 x^{2} = 32 - 16 x^{2} ★ C e d r i c j u n i o r$ $=> 24 x^{2} = 32 => x = \mp \sqrt{\frac{4}{3}} = \mp \frac{2 \sqrt{3}}{3}$ $S_{R} = {- \frac{2 \sqrt{3}}{3}; \frac{2 \sqrt{3}}{3}}$
	Answered by otchereabdullai last updated on 25/Feb/23

	$(\frac{x}{x - 2}) (\frac{x}{x - 2}) + (\frac{x}{x + 2}) (\frac{x}{x + 2}) = 2$ $\frac{x^{2}}{x (x - 2) - 2 (x - 2)} + \frac{x^{2}}{x (x + 2) + 2 (x + 2)} = 2$ $\frac{x^{2}}{x^{2} - 2 x - 2 x + 4} + \frac{x^{2}}{x^{2} + 2 x + 2 x + 4} = 2$ $\frac{x^{2}}{x^{2} - 4 x + 4} + \frac{x^{2}}{x^{2} + 4 x + 4} = 2$ $l c m = (x^{2} - 4 x + 4) (x^{2} + 4 x + 4)$ $m u l t i p l y t h r o u g h b y t h e l c m w e h a v e$ $x^{2} (x^{2} + 4 x + 4) + x^{2} (x^{2} - 4 x + 4) = 2 [(x^{2} - 4 x + 4) (x^{2} + 4 x + 4)]$ $e x p a n d a n d c a n c i l o p p o s i t e l i k e t e r m s$ $w e h a v e - 32 x^{2} + 8 x + 32 = 0$ $- 24 x^{2} + 32 = 0$ $- 24 x^{2} = - 32$ $24 x^{2} = 32$ $x^{2} = \frac{32}{24}$ $x = \pm \sqrt{\frac{32}{24}}$ $x = \pm \frac{4 \sqrt{2}}{2 \sqrt{6}}$ $x = \pm \frac{2 \sqrt{3}}{3}$
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