5. The equation $4 x-|3 x-| x+a||=9|x-1|$ has at least one root. Then the range of values for $a$ is $\qquad$ $\therefore$
5. $-8 \leqslant a \leqslant 6$.

Transform the original equation into
$$
9|x-1|+|3 x-| x+a||-4 x=0 \text {. }
$$

Consider the continuous function
$$
f(x)=9|x-1|+|3 x-| x+a||-4 x \text {. }
$$
(1) When $x \geqslant 1$, for any expansion of the absolute value, we have
$$
f(x)=9 x-9-4 x \pm 3 x \pm x \pm a=k x+m \text {, }
$$

where $k \geqslant 9-4-4=1>0$, so the function $f(x)$ is increasing.
(2) When $x<1$, similarly, we have
$$
f(x)=-9 x+9-4 x \pm 3 x \pm x \pm a=k x+m \text {, }
$$

where $k \leqslant-9-4+4=-9<0$, the function $f(x)$ is decreasing.

Therefore, $x=1$ is the minimum point of the function $f$, and the range of $f$ is $[f(1),+\infty)$. 
From the fact that the equation has a root, we know $f(1) \leqslant 0$, i.e.,
$$
\begin{array}{l}
|3-| 1+a|| \leqslant 4 \\
\Rightarrow-4 \leqslant|a+1|-3 \leqslant 4 \\
\Rightarrow|a+1| \leqslant 7 \\
\Rightarrow-8 \leqslant a \leqslant 6 .
\end{array}
$$