`{(x+xy+y=1),(y+yz+z=4),(z+zx+x=9):}`

Ta có:

$\bullet$ `x  + xy + y = 1 <=> (x+1)(y+1) - 1 = 1`

     `<=> (x+1)(y+1) = 2`   `\bb ((1))`

Tương tự: $\bullet$ `y + yz + z = 4 <=> (y+1)(z+1) = 5`   `\bb ((2))`

                  $\bullet$ `z + zx + x = 9 <=> (z+1)(x+1) = 10`   `\bb ((3))`

Lấy `\bb ((1))`, `\bb ((2))`, `\bb ((3))` nhân vế theo vế, ta được:

     `(x+1)(y+1) . (y+1)(z+1) . (z+1)(x+1) = 2 . 5 . 10`

`<=> [(x+1)(y+1)(z+1)]^2 = 100`

`<=>`\(\left[ \begin{array}{l}(x+1)(y+1)(z+1)=10\\(x+1)(y+1)(z+1)=-10\end{array} \right.\) 

$\color{red}\bullet$ `\bb \color{red}(TH_1: )` `(x+1)(y+1)(z+1)=10`   $(*)$

Thay `\bb ((1))`, `\bb ((2))`, `\bb ((3))` vào $(*)$. Khi đó: 

   `{((x+1)(y+1)=2),((y+1)(z+1) = 5),((z+1)(x+1) = 10):} => {(z+1=5),(x+1=2),(y+1=1):} <=> {(x=1),(y=0),(z=4):}`

$\color{red}\bullet$ `\bb \color{red}(TH_2: )` $(x+1)(y+1)(z+1)=-10$   $(*)$

Thay `\bb ((1))`, `\bb ((2))`, `\bb ((3))` vào $(*)$. Khi đó: 

   `{((x+1)(y+1)=2),((y+1)(z+1) = 5),((z+1)(x+1) = 10):} => {(z+1=-5),(x+1=-2),(y+1=-1):} <=>`$\begin{cases} x=-3\\y=-2\\z=-6 \end{cases}$

Vậy, `(x ; y ; z) \in {(1 ; 0 ; 4) ; `$(-3 ; -2 ; -6)$`}`