Solve in $ \mathbb{R} $ the equation $ [x]^5+\{ x\}^5 =x^5, $ where $ [],\{\} $ are the integer part, respectively, the fractional part.
1. Let \( x = k + t \) where \( k \) is an integer and \( 0 \leq t < 1 \). Then, the integer part of \( x \) is \( k \) and the fractional part of \( x \) is \( t \). Therefore, the given equation can be rewritten as:
   \[
   [x]^5 + \{x\}^5 = x^5 \implies k^5 + t^5 = (k + t)^5
   \]

2. Expanding \( (k + t)^5 \) using the binomial theorem, we get:
   \[
   (k + t)^5 = k^5 + 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4 + t^5
   \]

3. Substituting this back into the equation, we have:
   \[
   k^5 + t^5 = k^5 + 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4 + t^5
   \]

4. Simplifying, we get:
   \[
   0 = 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4
   \]

5. Factoring out \( t \) from the right-hand side, we obtain:
   \[
   0 = t(5k^4 + 10k^3t + 10k^2t^2 + 5kt^3)
   \]

6. This implies:
   \[
   t = 0 \quad \text{or} \quad 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0
   \]

7. If \( t = 0 \), then \( x = k \), which means \( x \) is an integer. Therefore, \( x \in \mathbb{Z} \).

8. If \( t \neq 0 \), then we have:
   \[
   5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0
   \]

9. Dividing the entire equation by \( t \), we get:
   \[
   5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0 \implies 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0
   \]

10. Factoring out \( 5k \) from the equation, we get:
    \[
    k(5k^3 + 10k^2t + 10kt^2 + 5t^3) = 0
    \]

11. This implies:
    \[
    k = 0 \quad \text{or} \quad 5k^3 + 10k^2t + 10kt^2 + 5t^3 = 0
    \]

12. If \( k = 0 \), then \( x = t \) where \( 0 \leq t < 1 \). Therefore, \( x \in [0, 1) \).

13. If \( k \neq 0 \), then we have:
    \[
    5k^3 + 10k^2t + 10kt^2 + 5t^3 = 0
    \]

14. Dividing the entire equation by 5, we get:
    \[
    k^3 + 2k^2t + 2kt^2 + t^3 = 0
    \]

15. Factoring the left-hand side, we get:
    \[
    (k + t)(k^2 + kt + t^2) = 0
    \]

16. This implies:
    \[
    k + t = 0 \quad \text{or} \quad k^2 + kt + t^2 = 0
    \]

17. If \( k + t = 0 \), then \( k = -t \). Since \( k \) is an integer and \( 0 \leq t < 1 \), \( k \) must be 0. Therefore, \( x = t \) where \( 0 \leq t < 1 \). Hence, \( x \in [0, 1) \).

18. If \( k^2 + kt + t^2 = 0 \), then since \( k \) and \( t \) are real numbers, this equation has no real solutions for \( k \neq 0 \) and \( 0 \leq t < 1 \).

19. Therefore, the only solutions are \( x \in [0, 1) \cup \mathbb{Z} \).

The final answer is \( \boxed{ [0, 1) \cup \mathbb{Z} } \).