$f(x)$ is a decreasing function defined on $(0,+\infty)$, if $f(2a^2+a+1)<f(3a^2-4a+1)$, then the range value of $a$ is________.
1. Given that \( f(x) \) is a decreasing function on \( (0, +\infty) \), and we have the inequality \( f(2a^2 + a + 1) < f(3a^2 - 4a + 1) \). Since \( f \) is decreasing, this implies:
   \[
   2a^2 + a + 1 > 3a^2 - 4a + 1
   \]

2. Simplify the inequality:
   \[
   2a^2 + a + 1 > 3a^2 - 4a + 1
   \]
   \[
   2a^2 + a + 1 - 3a^2 + 4a - 1 > 0
   \]
   \[
   -a^2 + 5a > 0
   \]
   \[
   a(-a + 5) > 0
   \]

3. Solve the inequality \( a(-a + 5) > 0 \):
   - The roots of the quadratic equation \( a(-a + 5) = 0 \) are \( a = 0 \) and \( a = 5 \).
   - The quadratic \( a(-a + 5) \) is a downward-opening parabola. Therefore, the inequality \( a(-a + 5) > 0 \) holds for \( 0 < a < 5 \).

4. Next, we need to ensure that both \( 2a^2 + a + 1 > 0 \) and \( 3a^2 - 4a + 1 > 0 \) for the values of \( a \) in the interval \( (0, 5) \).

5. Check the discriminant of \( 3a^2 - 4a + 1 \):
   \[
   \Delta = (-4)^2 - 4 \cdot 3 \cdot 1 = 16 - 12 = 4
   \]
   The discriminant is positive, so the quadratic \( 3a^2 - 4a + 1 \) has two real roots:
   \[
   a = \frac{4 \pm \sqrt{4}}{2 \cdot 3} = \frac{4 \pm 2}{6}
   \]
   \[
   a = 1 \quad \text{and} \quad a = \frac{1}{3}
   \]

6. The quadratic \( 3a^2 - 4a + 1 \) is positive outside the interval \( \left( \frac{1}{3}, 1 \right) \). Therefore, \( 3a^2 - 4a + 1 > 0 \) for \( a < \frac{1}{3} \) or \( a > 1 \).

7. Combine the intervals:
   - From step 3, we have \( 0 < a < 5 \).
   - From step 6, we have \( a < \frac{1}{3} \) or \( a > 1 \).

8. The intersection of these intervals is:
   \[
   0 < a < \frac{1}{3} \quad \text{or} \quad 1 < a < 5
   \]

The final answer is \( \boxed{ 0 < a < \frac{1}{3} } \) or \( 1 < a < 5 \).