 Let be $f: (0,+\infty)\rightarrow \mathbb{R}$ monoton-decreasing . 

If $f(2a^2+a+1)<f(3a^2-4a+1)$ find interval of $a$ .

1. Given that \( f \) is a monotonically decreasing function, we have the inequality:
   \[
   f(2a^2 + a + 1) < f(3a^2 - 4a + 1)
   \]
   This implies:
   \[
   2a^2 + a + 1 > 3a^2 - 4a + 1
   \]
   because for a monotonically decreasing function, if \( x < y \), then \( f(x) > f(y) \).

2. Simplify the inequality:
   \[
   2a^2 + a + 1 > 3a^2 - 4a + 1
   \]
   Subtract \( 3a^2 - 4a + 1 \) from both sides:
   \[
   2a^2 + a + 1 - (3a^2 - 4a + 1) > 0
   \]
   Simplify the expression:
   \[
   2a^2 + a + 1 - 3a^2 + 4a - 1 > 0
   \]
   \[
   -a^2 + 5a > 0
   \]
   Factor the quadratic expression:
   \[
   a(-a + 5) > 0
   \]
   This can be rewritten as:
   \[
   a(5 - a) > 0
   \]
   The solutions to this inequality are:
   \[
   0 < a < 5
   \]

3. Next, we need to consider the domain constraints for \( 2a^2 + a + 1 \) and \( 3a^2 - 4a + 1 \):
   - For \( 2a^2 + a + 1 > 0 \):
     \[
     2a^2 + a + 1 > 0
     \]
     Since the quadratic \( 2a^2 + a + 1 \) has a positive leading coefficient and its discriminant \( \Delta = b^2 - 4ac = 1 - 8 = -7 \) is negative, it is always positive for all \( a \in \mathbb{R} \).

   - For \( 3a^2 - 4a + 1 > 0 \):
     \[
     3a^2 - 4a + 1 > 0
     \]
     The roots of the quadratic equation \( 3a^2 - 4a + 1 = 0 \) are:
     \[
     a = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} = 1, \frac{1}{3}
     \]
     The quadratic \( 3a^2 - 4a + 1 \) is positive outside the interval \( \left( \frac{1}{3}, 1 \right) \):
     \[
     a \in (-\infty, \frac{1}{3}) \cup (1, +\infty)
     \]

4. Combine the intervals:
   \[
   a \in (0, 5) \cap (-\infty, +\infty) \cap \left( (-\infty, \frac{1}{3}) \cup (1, +\infty) \right)
   \]
   Simplify the intersection:
   \[
   a \in (0, 5) \cap \left( (-\infty, \frac{1}{3}) \cup (1, +\infty) \right)
   \]
   \[
   a \in (0, \frac{1}{3}) \cup (1, 5)
   \]

The final answer is \(\boxed{a \in (0, \frac{1}{3}) \cup (1, 5)}\)