6. The function $f(x)$ defined on $\mathbf{R}$ satisfies: when $x \in[0,1)$, $f(x)=2^{x}-x$, and for any real number $x$, $f(x)+f(x+1)=1$. Let $a=\log _{2} 3$, then the value of the expression $f(a)+f(2 a)+f(3 a)$ is $\qquad$ .
Solve:
$f(x+2)=1-f(x+1)=1-[1-f(x)]=f(x) \Rightarrow f(x)$ is a periodic function with a period of 2. Since $a=\log _{2} 3 \in(1,2), 2 a=\log _{2} 9 \in(3,4), 3 a=\log _{2} 27 \in(4,5)$,
and $f(a-1)=\frac{3}{2}-\log _{2} \frac{3}{2}, f(2 a-3)=\frac{9}{8}-\log _{2} \frac{9}{8}, f(3 a-4)=\frac{27}{16}-\log _{2} \frac{27}{16}$,
thus $f(a)=\log _{2} \frac{3}{2}-\frac{1}{2}, f(2 a)=\log _{2} \frac{9}{8}-\frac{1}{8}, f(3 a)=\frac{27}{16}-\log _{2} \frac{27}{16}$
$\Rightarrow f(a)+f(2 a)+f(3 a)=\frac{27}{16}-\frac{1}{2}-\frac{1}{8}+\log _{2}\left(\frac{3}{2} \cdot \frac{9}{8} \cdot \frac{16}{27}\right)=\frac{17}{16}$.