ขอบคุณครับ
ขอนำมาแปะในเว็บนี้ละกัน เพื่อป้องกันการสูญหาย
Obviously neither variable can be zero. Multiplying the second equation by $x$ and subtracting from the first we get
$$x-yzx^2=12-21x\iff 22x-12=yzx^2\iff xyz=22-\frac{12}{x}$$
Multiplying the third by $y$ and subtracting from the second, and multiplying the first by $z$ and subtracting from the third, we similarly find
$$xyz=31-\frac{21}{y},xyz=13-\frac{30}{z}$$
Hence $\displaystyle{22-\frac{12}{x}=31-\frac{21}{y}=13-\frac{30}{z}}$, which yields
$$y=\frac{7x}{4+3x}, z=\frac{10x}{4-3x}\qquad (*)$$
Now the first equation becomes
$$x+\frac{7x^2}{4+3x}+\frac{70x^3}{16-9x^2}=12$$
or, after simplifying,
$$5x^3+17x^2+2x-24=0$$
This factors to $(x-1)(x+2)(5x+12)=0$
For $x=1$ from $(*)$ we get $y=1, z=10$
For $x=-2$ we get $y=7, z=-2$
For $x=-\frac{12}{5}$ we get $y=\frac{21}{4}, z=-\frac{15}{7}$
$$\huge{(x,y,z)=(1,1,10),(-2,7,-2),(-\frac{12}{5},\frac{21}{4},-\frac{15}{7})}$$
