5. Given that $x$ is a real number, and
$$
|3 x-1|+|4 x-1|+\cdots+|17 x-1|
$$

is a definite constant. Then this constant is ( ).
(A) 75
(B) 15
(C) 10
(D) 5
5. D.

From the problem, we know that after removing the absolute value symbols, $x$ should be eliminated.
And $3+4+\cdots+12=13+14+\cdots+17$, so the original expression is
$$
\begin{aligned}
= & (1-3 x)+(1-4 x)+\cdots+(1-12 x)+ \\
& (13 x-1)+(14 x-1)+\cdots+(17 x-1) \\
= & 5 .
\end{aligned}
$$