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
$${with}\:\:{end}\:{terme}..\:+\left(\mathrm{2020}^{\mathrm{2}} −\mathrm{2021}^{\mathrm{2}} −\mathrm{2022}^{\mathrm{2}} +\mathrm{2023}^{\mathrm{2}} \right) \\ $$$${Resultat}:\:\mathrm{2020}+\mathrm{4}\:=\mathrm{2024} \\ $$
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))}](Q180699.png)
$$\mathrm{Probleme}\:\mathrm{5} \\ $$$$\begin{cases}{\mathrm{xy}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{3p}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{xy}−\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\mathrm{with}\:\:\:\mathrm{y}\neq\mathrm{0} \\ $$$$\:\:\:\:\frac{\left(\mathrm{1}\right)}{\left(\mathrm{2}\right)}\Rightarrow\frac{\mathrm{xy}+\frac{\mathrm{x}}{\mathrm{y}}}{\mathrm{xy}−\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{3}\:\Rightarrow\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{3}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\mathrm{y}=\pm\sqrt{\mathrm{2}} \\ $$$$\left.\:\:\mathrm{1}\right)\mathrm{y}=−\sqrt{\mathrm{2}}\:\:\: \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2p}}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{x}\sqrt{\mathrm{2}}}{\mathrm{2p}}+\mathrm{2}\:\:\:\:\:\:\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2p}^{\mathrm{2}} }−\mathrm{8} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{16}{p}^{\mathrm{2}} }\:\right)}{\mathrm{4}{p}} \\ $$$$\left.\:\mathrm{2}\right){y}=\sqrt{\mathrm{2}} \\ $$$$\:\:\:\mathrm{x}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{x}=\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\:\: \\ $$$$\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{x}\sqrt{\mathrm{2}}}{\mathrm{2p}}+\mathrm{2}\:\:\:\:\mathrm{x}_{\mathrm{2}} =−\mathrm{x}_{\mathrm{1}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}<\mathrm{p}<\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\left(\mathrm{2}\:{racines}\:\:{reels}\right) \\ $$$$\left.\mathrm{p}\left.\in\right]−\infty,\frac{−\mathrm{1}}{\mathrm{4}}\right]\cup\left[\frac{\mathrm{1}}{\mathrm{4}},+\infty\left[\:\:\left(\mathrm{2}\:\:{racines}\:{coppmplexes}\right)\right.\right. \\ $$$${Conclusion} \\ $$$${S}=\left\{\left(\frac{\sqrt{\mathrm{2}\left(\right.}\:\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{16}{p}^{\mathrm{2}} }}{\mathrm{2}{p}},−\sqrt{\mathrm{2}}\:\right);\left(−\frac{\sqrt{\mathrm{2}}\:\left(\pm\mathrm{1}\sqrt{\mathrm{1}−\mathrm{16}{p}^{\mathrm{2}} }\right.}{\mathrm{4}{p}},+\sqrt{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$
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}](Q180700.png)
$$\mathrm{probleme}\:\mathrm{5}\left(\mathrm{partie}\:\mathrm{2}\right) \\ $$$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:\left(−\right)\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}} \\ $$$$\mathrm{si}\:\:\:\mathrm{signe}\:−\mathrm{alors}\:\: \\ $$$$\mathrm{0}=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}}\:\Rightarrow\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}=\mathrm{0} \\ $$$$\mathrm{x}=\varnothing\left({pas}\:{de}\:{solutions}\right) \\ $$$${signe}\left(+\right) \\ $$$$\mathrm{2}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}\:=\frac{\mathrm{3}}{\mathrm{2}}×\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}}\:\: \\ $$$$\mathrm{4}\left(\mathrm{x}+\sqrt{\mathrm{x}}\right)=\frac{\mathrm{9}}{\mathrm{4}}×\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}\:\: \\ $$$$\mathrm{16}\left(\mathrm{x}+\sqrt{\mathrm{x}}\:\right)^{\mathrm{2}} −\mathrm{9x}=\mathrm{0} \\ $$$$\mathrm{16x}^{\mathrm{2}} +\mathrm{32x}\sqrt{\mathrm{x}}\:+\mathrm{7x}=\mathrm{0} \\ $$$$\mathrm{x}+\mathrm{32}\sqrt{\mathrm{x}}\:+\mathrm{7}=\mathrm{0}\:\:\:\:\left(\mathrm{avec}\:\mathrm{x}\neq\mathrm{0}\right) \\ $$$$\:\:\sqrt{\mathrm{x}}\:=\mathrm{t}\:\:\:\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{32t}+\mathrm{7}=\mathrm{0} \\ $$$$\:\:\mathrm{16}^{\mathrm{2}} −\mathrm{7}=\left[\mathrm{15},\mathrm{78}\right]^{\mathrm{2}} \:\:\:\: \\ $$$$\:\:\mathrm{x}=−\mathrm{16}\pm\sqrt{\mathrm{249}}\:=−\mathrm{16}\pm\sqrt{\mathrm{3}×\mathrm{83}}\:\cong−\mathrm{16}\pm\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{x}=\left\{\mathrm{31},\mathrm{78}{i}\:\:;\:\:\mathrm{0},\mathrm{22}{i}\right\} \\ $$$$ \\ $$$$ \\ $$