$\begin{cases} (y-1)(x-1)=6\\(y-1)(z-1)=12\\(x-1)(z-1)=8 \end{cases}$
$⇔$$\begin{cases} y-1=\dfrac{6}{x-1}\\\dfrac{6}{x-1}(z-1)=12\\(z-1)=\dfrac{8}{x-1} \end{cases}$
$⇔$$\begin{cases} y-1=\dfrac{6}{x-1}\\\dfrac{6}{x-1}.\dfrac{8}{x-1}=12\\(z-1)=\dfrac{8}{x-1} \end{cases}$
$⇔$$\begin{cases} y-1=\dfrac{6}{x-1}\\\dfrac{48}{x^2-2x+1}=12\\(z-1)=\dfrac{8}{x-1} \end{cases}$
$⇔$$\begin{cases} y-1=\dfrac{6}{x-1}\\48=12x^2-24x+12\\(z-1)=\dfrac{8}{x-1} \end{cases}$
$⇔$$\begin{cases} y-1=\dfrac{6}{x-1}\\12x^2-24x-36=0\\(z-1)=\dfrac{8}{x-1} \end{cases}$
Nếu $x=3⇒y=\dfrac{6}{3-1}+1=4$
$⇒z=\dfrac{8}{3-1}+1=5$
$x=-1⇒y=\dfrac{6}{-1-1}+1=-2$
$⇒z=\dfrac{8}{-1-1}+1=-3$
