Functions

It is given that:

$\dfrac {1}{[x]} + \dfrac {1}{[2x]} = \{x\} + \dfrac{1}{3}$

$\Rightarrow \{x\} = \dfrac {1}{[x]} + \dfrac {1}{[2x]} - \dfrac{1}{3} = \begin{cases} \dfrac {1}{[x]} + \dfrac {1}{2[x]} - \dfrac{1}{3} & \{x\} < 0.5 \\ \dfrac {1}{[x]} + \dfrac {1}{2[x]+1} - \dfrac{1}{3} & \{x\} \ge 0.5 \end{cases}$

$=\begin{cases} \dfrac {9-2[x]}{6[x]} & \{x\} < 0.5 \\ \dfrac {3+8[x]-2[x]^2}{6[x]^2+3[x]} & \{x\} \ge 0.5 \end{cases}$

Since $\quad 0<\{x\}<1$ :

$\Rightarrow \begin{cases} 0< \dfrac {9-2[x]}{6[x]} <1 \\ 0 < \dfrac {3+8[x]-2[x]^2}{6[x]^2+3[x]} < 1 \end{cases} \Rightarrow 2\le [x] \le 4$

Let us check out the values of $\{x\}$ for $[x] = 2,3,4$ .

$\begin{array} {cccc} [x] & \frac {9-2[x]}{6[x]} & \frac {3+8[x]-2[x]^2}{6[x]^2+3[x]} & x = [x]+\{x\} \\ 2 & \dfrac{5}{12} & \color{#D61F06} {\dfrac{11}{30}<0.5} & \dfrac{29}{12} \\ 3 & \dfrac{1}{6} & \color{#D61F06} {\dfrac{1}{7}<0.5} & \dfrac{19}{6} \\ 4 & \dfrac{1}{24} & \color{#D61F06} {\dfrac{1}{36}<0.5} & \dfrac{97}{24} \end{array}$

Therefore, $x_1+x_2+x_3 = \dfrac{29}{12}+ \dfrac{19}{6} + \dfrac{97}{24} = \dfrac{231}{24} = \boxed{9.625}$