Let a sequences: $ x_0\in [0;1],x_{n\plus{}1}\equal{}\frac56\minus{}\frac43 \Big|x_n\minus{}\frac12\Big|$. Find the "best" $ |a;b|$ so that for all $ x_0$ we have $ x_{2009}\in [a;b]$
To solve the problem, we need to analyze the behavior of the sequence defined by \( x_{n+1} = \frac{5}{6} - \frac{4}{3} \left| x_n - \frac{1}{2} \right| \) and find the best interval \([a, b]\) such that \( x_{2009} \in [a, b] \) for any initial value \( x_0 \in [0, 1] \).

1. **Initial Analysis**:
   - The sequence starts with \( x_0 \in [0, 1] \).
   - The recursive formula is \( x_{n+1} = \frac{5}{6} - \frac{4}{3} \left| x_n - \frac{1}{2} \right| \).

2. **Behavior of the Sequence**:
   - Let's analyze the behavior of the function \( f(x) = \frac{5}{6} - \frac{4}{3} \left| x - \frac{1}{2} \right| \).
   - For \( x \in [0, 1] \):
     - If \( x \in [0, \frac{1}{2}] \), then \( \left| x - \frac{1}{2} \right| = \frac{1}{2} - x \).
     - If \( x \in [\frac{1}{2}, 1] \), then \( \left| x - \frac{1}{2} \right| = x - \frac{1}{2} \).

3. **Piecewise Function**:
   - For \( x \in [0, \frac{1}{2}] \):
     \[
     f(x) = \frac{5}{6} - \frac{4}{3} \left( \frac{1}{2} - x \right) = \frac{5}{6} - \frac{2}{3} + \frac{4}{3} x = \frac{1}{6} + \frac{4}{3} x
     \]
   - For \( x \in [\frac{1}{2}, 1] \):
     \[
     f(x) = \frac{5}{6} - \frac{4}{3} \left( x - \frac{1}{2} \right) = \frac{5}{6} - \frac{4}{3} x + \frac{2}{3} = \frac{7}{6} - \frac{4}{3} x
     \]

4. **Fixed Points and Convergence**:
   - To find the fixed points, solve \( f(x) = x \):
     - For \( x \in [0, \frac{1}{2}] \):
       \[
       \frac{1}{6} + \frac{4}{3} x = x \implies \frac{1}{6} = x - \frac{4}{3} x \implies \frac{1}{6} = -\frac{1}{3} x \implies x = -\frac{1}{2} \text{ (not in } [0, \frac{1}{2}])
       \]
     - For \( x \in [\frac{1}{2}, 1] \):
       \[
       \frac{7}{6} - \frac{4}{3} x = x \implies \frac{7}{6} = x + \frac{4}{3} x \implies \frac{7}{6} = \frac{7}{3} x \implies x = \frac{1}{2}
       \]

5. **Interval Analysis**:
   - We need to find the interval \([a, b]\) such that \( x_n \) remains within this interval for all \( n \geq 2 \).
   - From the given solution, it is shown that \( x_n \in \left[ \frac{7}{18}, \frac{5}{6} \right] \) for all \( n \geq 2 \).

6. **Verification**:
   - Let \( v_0 = \frac{7}{18} \) and \( v_{n+1} = \frac{9}{8} - \frac{3}{4} v_n \).
   - Verify that \( v_n \in \left[ \frac{1}{2}, \frac{5}{6} \right] \) implies \( v_{n+1} \in \left[ \frac{1}{2}, \frac{5}{6} \right] \).

7. **Conclusion**:
   - For \( x_0 = v_{2010} \in \left[ \frac{1}{2}, \frac{5}{6} \right] \), we have \( x_{2009} = v_1 = \frac{5}{6} \).
   - For \( x_0 = v_{2009} \in \left[ \frac{1}{2}, \frac{5}{6} \right] \), we have \( x_{2009} = v_0 = \frac{7}{18} \).

Thus, the best interval \([a, b]\) such that \( x_{2009} \in [a, b] \) for any \( x_0 \in [0, 1] \) is \(\left[ \frac{7}{18}, \frac{5}{6} \right]\).

The final answer is \( \boxed{ \left[ \frac{7}{18}, \frac{5}{6} \right] } \).