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Question Number 180667 by cortano1 last updated on 15/Nov/22

Commented by JDamian last updated on 15/Nov/22
I never expected this from you :(
	Answered by a.lgnaoui last updated on 15/Nov/22

	Commented by a.lgnaoui last updated on 15/Nov/22

	$w i t h e n d t e r m e . . + (2020^{2} - 2021^{2} - 2022^{2} + 2023^{2})$ $R e s u l t a t : 2020 + 4 = 2024$
	Answered by a.lgnaoui last updated on 15/Nov/22
	$Probleme 5 { ((xy+(x/y)=3p(x^2 +y^2 ) (1))),((xy−(x/y)=p(x^2 +y^2 ) (2))) :} with y≠0 (((1))/((2)))⇒((xy+(x/y))/(xy−(x/y)))=3 ⇒(y^2 +1)=3(y^2 −1) y=±(√2) 1)y=−(√2) x^2 +2+((√2)/(2p)) x=0 x^2 +((x(√2))/(2p))+2 △=(1/(2p^2 ))−8 x_1 =(((√2) (1±(√(1−16p^2 )) ))/(4p)) 2)y=(√2) x(√2) −((√2)/2)x=p(x^2 +2) x^2 −((x(√2))/(2p))+2 x_2 =−x_1 ((−1)/4)<p<(1/4) (2 racines reels) p∈]−∞,((−1)/4)]∪[(1/4),+∞[ (2 racines coppmplexes) Conclusion S={((((√(2()) 1±(√(1−16p^2 )))/(2p)),−(√2) );(−(((√2) (±1(√(1−16p^2 )))/(4p)),+(√2))}$
	$Probleme 5$ ${\begin{cases} xy + \frac{x}{y} = 3 p (x^{2} + y^{2}) (1) \\ xy - \frac{x}{y} = p (x^{2} + y^{2}) (2) \end{cases}$ $with y \neq 0$ $\frac{(1)}{(2)} \Rightarrow \frac{xy + \frac{x}{y}}{xy - \frac{x}{y}} = 3 \Rightarrow (y^{2} + 1) = 3 (y^{2} - 1)$ $y = \pm \sqrt{2}$ $1) y = - \sqrt{2}$ $x^{2} + 2 + \frac{\sqrt{2}}{2 p} x = 0$ $x^{2} + \frac{x \sqrt{2}}{2 p} + 2 △ = \frac{1}{{2 p}^{2}} - 8$ $x_{1} = \frac{\sqrt{2} (1 \pm \sqrt{1 - 16 p^{2}})}{4 p}$ $2) y = \sqrt{2}$ $x \sqrt{2} - \frac{\sqrt{2}}{2} x = p (x^{2} + 2)$ $x^{2} - \frac{x \sqrt{2}}{2 p} + 2 x_{2} = - x_{1}$ $\frac{- 1}{4} < p < \frac{1}{4} (2 r a c i n e s r e e l s)$ $p \in] - \infty, \frac{- 1}{4}] \cup [\frac{1}{4}, + \infty [(2 r a c i n e s c o p p m p l e x e s)$ $C o n c l u s i o n$ $S = {(\frac{\sqrt{2 (} 1 \pm \sqrt{1 - 16 p^{2}}}{2 p}, - \sqrt{2}); (- \frac{\sqrt{2} (\pm 1 \sqrt{1 - 16 p^{2}}}{4 p}, + \sqrt{2})}$
	Answered by a.lgnaoui last updated on 15/Nov/22
	$probleme 5(partie 2) (√(x+(√x) )) (−)(√(x+(√x) )) =(3/2)(√(x/(x+(√x)))) si signe −alors 0=(3/2)(√(x/(x+(√x)))) ⇒(x/(x+(√x)))=0 x=∅(pas de solutions) signe(+) 2(√(x+(√x))) =(3/2)×(√(x/(x+(√x)))) 4(x+(√x))=(9/4)×(x/(x+(√x))) 16(x+(√x) )^2 −9x=0 16x^2 +32x(√x) +7x=0 x+32(√x) +7=0 (avec x≠0) (√x) =t t^2 +32t+7=0 16^2 −7=[15,78]^2 x=−16±(√(249)) =−16±(√(3×83)) ≅−16±9(√3) x={31,78i ; 0,22i}$
	$probleme 5 (partie 2)$ $\sqrt{x + \sqrt{x}} (-) \sqrt{x + \sqrt{x}} = \frac{3}{2} \sqrt{\frac{x}{x + \sqrt{x}}}$ $si signe - alors$ $0 = \frac{3}{2} \sqrt{\frac{x}{x + \sqrt{x}}} \Rightarrow \frac{x}{x + \sqrt{x}} = 0$ $x = \emptyset (p a s d e s o l u t i o n s)$ $s i g n e (+)$ $2 \sqrt{x + \sqrt{x}} = \frac{3}{2} \times \sqrt{\frac{x}{x + \sqrt{x}}}$ $4 (x + \sqrt{x}) = \frac{9}{4} \times \frac{x}{x + \sqrt{x}}$ $16 {(x + \sqrt{x})}^{2} - 9 x = 0$ ${16 x}^{2} + 32 x \sqrt{x} + 7 x = 0$ $x + 32 \sqrt{x} + 7 = 0 (avec x \neq 0)$ $\sqrt{x} = t t^{2} + 32 t + 7 = 0$ $16^{2} - 7 = {[15, 78]}^{2}$ $x = - 16 \pm \sqrt{249} = - 16 \pm \sqrt{3 \times 83} ≅ - 16 \pm 9 \sqrt{3}$ $x = {31, 78 i; 0, 22 i}$
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