Question

2. Sabemos que o numero real C e números reais nao-nulos x, y e z, dois a dois distintos, satisfazem:
$\mathsf{x + \dfrac{y}{z} + \dfrac{z}{y} = y + \dfrac{x}{z} + \dfrac{z}{x} = z + \dfrac{x}{y} + \dfrac{y}{x} = C}$
a) Mostre que C = -1
b) Exiba ao menos uma solução (x,y,z) para a equação dada.

Answer 1

$\text{Primera ecuaci\'on: }\\ \\ x + \dfrac{y}{z} + \dfrac{z}{y} = y + \dfrac{x}{z} + \dfrac{z}{x}\\ \\ \\ x^2yz+xy^2+xz^2=xy^2z+x^2y+yz^2\\ \\ xyz(x-y)+xy(y-x)+z^2(x-y)=0\\ \\ (x-y)(xyz-xy+z^2)=0\\ \\ x=y~\vee~xyz-xy+z^2=0$


$\text{Segunda ecuaci\'on}\\ \\ x + \dfrac{y}{z} + \dfrac{z}{y} = z + \dfrac{x}{y} + \dfrac{y}{x} \\ \\ \\ x^2yz+xy^2+xz^2=xyz^2+x^2z+y^2z\\ \\ xyz(x-z)+y^2(x-z)+xz(z-x)=0\\ \\ (x-z)(xyz+y^2-xz)=0\\ \\ x=z~\vee~xyz+y^2-xz=0$


$\text{Tercera ecuaci\'on}\\ \\ y + \dfrac{x}{z} + \dfrac{z}{x} = z + \dfrac{x}{y} + \dfrac{y}{x}\\ \\ \\ xy^2z+x^2y+yz^2=xyz^2+x^2z+y^2z\\ \\ xyz(y-z)+x^2(y-z)+yz(z-y)=0\\ \\ (y-z)(xyz+x^2-yz)=0\\ \\ y=z~\vee~xyz+x^2-yz=0$


$\text{Como }x\neq y\neq z\text{ entonces}\\ \\ S=\begin{cases} xyz-xy+z^2=0\\ xyz-xz+y^2=0\\ xyz-yz+x^2=0 \end{cases}\\ \\ \\ E_1:~z^2-xy=y^2-xz=x^2-yz\\ \\ z^2-xy=y^2-xz\\ x(z-y)=y^2-z^2\\ x=-(y+z)\\ \boxed{x+y+z=0}\\ \\ S_1: xyz-xy+z^2=0\\ \\ -(y+z)yz+(y+z)y+z^2=0\\ -y^2z-yz^2+y^2+zy+z^2=0\\ z^2(1-y)+(y-y^2)z+y^2=0\\ \\ z=\dfrac{y^2-y\pm\sqrt{(y^2-y)^2-4y^2(1-y)}}{2(1-y)}$


$z=\dfrac{y^2-y\pm y\sqrt{(y-1)(y+3)}}{2(1-y)}=-\dfrac{y}{2}\pm \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}\\ \\ \\ x=-\dfrac{y}{2}\mp \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}\\ \\ \\ \text{Indistintamente podemos elegir}\\ \\ z=-\dfrac{y}{2}+ \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}\\ \\ \\ x=-\dfrac{y}{2}- \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}$


Luego

$C=x+\dfrac{y}{z}+\dfrac{z}{y}\\ \\ \\ C=-\dfrac{y}{2}- \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}+\dfrac{y}{-\dfrac{1}{2}y+ \dfrac{1}{2}y\sqrt{\dfrac{y+3}{y-1}}}+\dfrac{-\dfrac{y}{2}+ \dfrac{y}{2}\sqrt{\dfrac{y+3}{y-1}}}{y}\\ \\ \\ C=-\dfrac{y}{2}-\dfrac{y}{2}\sqrt{\dfrac{y+3}{y-1}}+\dfrac{2}{-1+ \sqrt{\dfrac{y+3}{y-1}}}-\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{y+3}{y-1}}$


$C=-\left(\dfrac{y+1}{2}\right)- \left(\dfrac{y-1}{2}\right)\sqrt{\dfrac{y+3}{y-1}}+\dfrac{y-1}{2}+\dfrac{y-1}{2}\sqrt{\dfrac{y+3}{y-1}}\\ \\ \\ \boxed{C=-1}\\ \\ \\ \text{La soluci\'on del sistema anterior era: }\\ \\ \\ (x,y,z)=\left(-\dfrac{y}{2}- \dfrac{y}{2}\sqrt{\dfrac{y+3}{y-1}}~,~y~,~-\dfrac{y}{2}+ \dfrac{y}{2}\sqrt{\dfrac{y+3}{y-1}}\right)\\ \\ \\ \text{Una soluci\'on puede ser haciendo }y=2\\ \\ (x,y,z)=\left(-1-\sqrt{5},2,-1+\sqrt{5}~\right)$