8. (40 points) Find all ordered triples $(x, y, z)$, such that $x, y, z \in Q_{+}$, and $x+\frac{1}{y}$, $y+\frac{1}{z}$, $z+\frac{1}{x}$ are all integers.
8. Notice that
$$
\begin{array}{l}
\left(x+\frac{1}{y}\right)\left(y+\frac{1}{z}\right)\left(z+\frac{1}{x}\right) \\
=x y z+\frac{1}{x y z}+x+\frac{1}{y}+y+\frac{1}{z}+z+\frac{1}{x} .
\end{array}
$$

Thus, $h=x y z+\frac{1}{x y z}$ is an integer.
Then $x y z=\frac{h \pm \sqrt{h^{2}-4}}{2}$.
Since the difference between two perfect squares that differ by 4 can only be 0 and 4, we know $h=2, x y z=1$.

If two of $x, y, z$ are 1, then the third must also be 1. Hence, $(x, y, z)=(1,1,1)$.

If only one of $x, y, z$ is 1, assume $z=1$. Then $y+1$ and $1+\frac{1}{x}$ are both integers. Thus,
$y=m, x=\frac{1}{n}\left(m, n \in \mathbf{N}_{+}\right)$.
Also, $x+\frac{1}{y}=\frac{1}{n}+\frac{1}{m} \in \mathbf{N}_{+} \Rightarrow m=n=2$.
Hence, $(x, y, z)=\left(\frac{1}{2}, 2,1\right)$ and its permutations.
If none of $x, y, z$ are 1, assume $x>1$ and $z>1$, this contradicts $x>1$. So, $k=1, x=\frac{q}{q-p}$.
Thus, $y=\frac{1}{x z}=\frac{q-p}{q} \times \frac{q}{p}=\frac{q-p}{p}$.
By $\left.y+\frac{1}{z}=\frac{2 q-p}{p} \in \mathbf{N}_{+} \Rightarrow p \right\rvert\, 2 q \Rightarrow p \leqslant 2$.
Also, $\left.x+\frac{1}{y}=1+\frac{2 p}{q-p} \in \mathbf{N}_{+} \Rightarrow(q-p) \right\rvert\, 2 p$.
When $p=1$, $q-p=2$, thus, $q=3$;
When $p=2$, $q-p=1$, thus, $q=3$.
Hence, $(x, y, z)=\left(\frac{3}{2}, 2, \frac{1}{3}\right),\left(3, \frac{1}{2}, \frac{2}{3}\right)$ and their permutations.