6. $[x]$ represents the greatest integer not greater than $x$, the equation
$$
[2 x]+[3 x]=8 x-\frac{7}{2}
$$

all real solutions are $\qquad$ .
6. $\frac{13}{16}, \frac{17}{16}$.

From $2 x-1<[2 x] \leqslant 2 x, 3 x-1<[3 x] \leqslant 3 x$ and the given equation, we get
$$
\begin{array}{l}
5 x-2<8 x-\frac{7}{2} \leqslant 5 x \Leftrightarrow \frac{1}{2}<x \leqslant \frac{7}{6} \\
\Leftrightarrow \frac{1}{2}<8 x-\frac{7}{2} \leqslant \frac{35}{6} .
\end{array}
$$

Since $8 x-\frac{7}{2}=[2 x]+[3 x]$ is an integer, we have
$$
8 x-\frac{7}{2}=1,2,3,4,5 \text {. }
$$

Solving for $x$ gives $x=\frac{9}{16}, \frac{11}{16}, \frac{13}{16}, \frac{15}{16}, \frac{17}{16}$.
Upon verification, only $x=\frac{13}{16}, \frac{17}{16}$ are solutions to the given equation.