6. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$. Then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$ .
6.4.

Definition: $f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$.
It can be rewritten as $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\right)^{x}+\left(\frac{5}{6}\right)^{x}$.
It is easy to see that the function $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\right)^{x}+\left(\frac{5}{6}\right)^{x}$
is a decreasing function defined on the set of real numbers.
Furthermore, since $f(3)=1, f(1)=2, f(0)=3$, the solutions to the equation $(t-1)(t-2)(t-3)=0$ are $0, 1, 3$.