Jolie .

Trouvez tous les triplets (x,y,z) de IR tel que :

$Jolie . Gif$

alidos a écrit:: Trouvez tous les triplets (x,y,z) de IR tel que :

$Jolie . Gif$

Personne

Expert grade2 Nombre de messages : 310 Age : 27 Date d'inscription : 10/10/2012

Bon je vais travailler avec a,b,c au-lieu de x,y,z

$Jolie . Gif.latex?\\%20ab+bc+ac=1%20\implies(a+b)(b+c)=b^2+1%20\\%20ab+bc+ac=1%20\implies(a+c)(a+b)=a^2+1%20~~~~~~~~~~(1)%20\\%20ab+bc+ac=1%20\implies(a+c)(b+c)=c^2+1%20\\%205%20\left%20(%20a+\frac{1}{a}%20\right)=12\left%20(b+\frac{1}{b}%20\right)=13%20\left%20(c+\frac{1}{c}%20\right)%20\\%20\\%20\Longleftrightarrow%205b%20\left%20(a^2+1%20\right)=12%20\left%20(b^2+1%20\right)%20\%20\wedge%20\%2012c(b^2+1)=%2013b%20\left(%20c^2+1%20\right%20)(2)%20\\%20\text{De%20(1)%20et%20(2)%20on%20a%20:}%20\\%20(2)\Longleftrightarrow%2017bc+5ab=12%20\%20\wedge%20\%2025ac+12bc=13%20\%20(%20\wedge%20\%20ab+bc+ac=1)%20\\%20\text{En%20posant%20x=ab,%20y=bc,%20z=ac%20on%20trouve%20:%20}%20\left\{\begin{matrix}%205x+17y=12%20&%20\\%2012y+25z=13%20&%20\\%20x+y+z=1%20&%20\end{matrix}\right.%20\\\\%20\text{Syst%C3%A8me%20assez%20facile.%20En%20le%20r%C3%A9solvant%20on%20trouve%20:%20}%20\\%20\left\{\begin{matrix}%20x=\frac{2}{15}%20&%20\\%20y=\frac{2}{3}%20&%20\\%20z=%20\frac{1}{5}%20&%20\end{matrix}\right.\Longleftrightarrow%20\left\{\begin{matrix}%20ab=\frac{2}{15}%20&%20\\%20bc=\frac{2}{3}%20&%20\\%20ac=%20\frac{1}{5}%20&%20\end{matrix}\right.%20\Longleftrightarrow%20\left\{\begin{matrix}%20a=\frac{1}{5}%20&%20\\%20b=\frac{2}{3}%20&%20\\%20c=1%20&%20\end{matrix}\right$

$Jolie . Gif$

S={(1/5 , 2/3 ,1) et (-1/5,-2/3,-1)

Bref Good Mr humber ^^

» aussi jolie jolie jolie!!!
» jolie...
» jolie !
» jolie jolie jolie!!!
» Nice inequality