5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$.
5.2.

For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0\right),\left(x_{2}, 0\right) \text { and } \\
12,
\end{array}
$$

as shown in Figure 6.
The solution to the inequality is

the set of $x$ values for which both functions are either both positive or both negative,
thus the solution set is $\left(1, x_{1}\right] ;\left(2, x_{2}\right]$.
The total length $L$ of the solution set is
$$
L=\left(x_{1}-1\right)+\left(x_{2}-2\right)=\left(x_{1}+x_{2}\right)-(1+2) \text {. }
$$
Given $x_{1}+x_{2}=1+2+\frac{2}{3}(1+2)$,
$$
\therefore L=\frac{2}{3}(1+2)=2 \text {. }
$$