Giải:
Theo đề bài ta có:
\(\left(x+1\right).yz=A+1\)
\(\Rightarrow xyz+yz=A+1\)
\(\Rightarrow A+yz=A+1\Rightarrow yz=1\left(1\right)\)
\(\left(y+2\right).xz=A+2\)
\(\Rightarrow xyz+xz=A+2\)
\(\Rightarrow A+2xz=A+2\Rightarrow xz=1\left(2\right)\)
\(\left(z+2\right).xy=A+8\)
\(\Rightarrow xyz+xy=A+8\)
\(\Rightarrow A+2xy=A+8\Rightarrow xy=4\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow yzxzxy=1.1.4\)
\(\Rightarrow x^2y^2z^2=4\)
\(\Rightarrow\left(xyz\right)^2=4\)
\(\Rightarrow xyz=\sqrt{4}\)
\(\Rightarrow xyz=2\)
Vậy \(xyz=2\)