Ta có: `xy . yz . xz = 1/3 . (-2/5) . (-3/10)`

`=> xyyzxz = 1/25`

`=>x^2y^2z^2 =1/25`

`=> (xyz)^2 = 1/25`

`=>` \(\left[ \begin{array}{l}xyz = \frac{1}{5}\\xyz= -\frac{1}{5}\end{array} \right.\)

+) Nếu `xyz= 1/5`

Mà `xy = 1/3`

`=> 1/3 z= 1/5`

`=> z= 1/5 : 1/3`

`=> z= 1/5 .3`

`=> z= 3/5`

`=> y= -2/5 : 3/5 = -2/5 . 5/3 = -2/3`

`=> x = 1/3 : (-2/3) = 1/3 . (-3/2) = -1/2`

+) Nếu `xyz= -1/5`

Mà `xy= 1/3`

`=> 1/3 z = -1/5`

`=> z= -1/5 : 1/3`

`=> z= -1/5. 3`

`=> z= -3/5`

`=> y= -2/5 : (-3/5)`

`=> y= -2/5 . (-5/3)`

`=>y= 2/3`

`=> x= 1/3 : 2/3`

`=> x= 1/3 . 3/2 = 1/2`

Vậy `x = -1/2; y= -2/3 ; z= 3/5` hoặc `x= 1/2 ; y =2/3 ; z= -3/5`

 Theo đề ra , ta có : 
`x.y.y.z.x.z=1/3 . (-2/5).(-3/10)`
`=>.y.y.z.x.z=1/25`
`=>(x.y.z)²=1/25`
`=>`  \(\left[ \begin{array}{l}x.y.z=\dfrac{1}{5}\\x.y.z=-\dfrac{1}{5}\end{array} \right.\) 
+ Nếu `x.y.z=1/5` thì :
`=>`  \(\left[ \begin{array}{l}x=\dfrac{1}{5}:\dfrac{1}{3}\\y=\dfrac{1}{5}:(-\dfrac{2}{5})\\z=\dfrac{1}{5}:(-\dfrac{3}{10})\end{array} \right.\) 
`=>` \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\y=-\dfrac{1}{2}\\z=-\dfrac{2}{3}\end{array} \right.\) 
+ Nếu `x.y.z=-1/5` thì :
`=>`  \(\left[ \begin{array}{l}x=(-\dfrac{1}{5}):\dfrac{1}{3}\\y=(-\dfrac{1}{5}):(-\dfrac{2}{5})\\z=(-\dfrac{1}{5}):(-\dfrac{3}{10})\end{array} \right.\) 
`=>` \(\left[ \begin{array}{l}x=-\dfrac{3}{5}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{array} \right.\) 
Vậy `(x;y;z)=(3/5:-1/2;-2/3),(-3/5;1/2;2/3)`