Let `x=(5+2sqrt6)^(n),ninN`, then find the value of `x-x^(2)+x[x]`, where [.] denotes greatest integer function.

Answer

To solve the problem, we need to find the value of \( x - x^2 + x \lfloor x \rfloor \), where \( x = (5 + 2\sqrt{6})^n \) and \( n \) is a natural number. Here, \( \lfloor x \rfloor \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Define \( x \) and \( y \)**:
Let \( x = (5 + 2\sqrt{6})^n \) and \( y = (5 - 2\sqrt{6})^n \). Note that \( 5 - 2\sqrt{6} \) is a small positive number less than 1, so \( y \) will be a very small positive number as \( n \) increases.

2. **Find \( x + y \)**:
We can express \( x + y \):
\[ x + y = (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n \]
This sum will be an integer because it can be shown that the terms involving \( \sqrt{6} \) cancel out.

3. **Find \( \lfloor x \rfloor \)**:
Since \( y \) is very small, we can express \( x \) in terms of its integer and fractional parts:
\[ x = \lfloor x \rfloor + \{x\} \]
where \( \{x\} = x - \lfloor x \rfloor \) is the fractional part of \( x \).

4. **Relate \( y \) to \( \{x\} \)**:
We can express \( y \) as:
\[ y = 1 - \{x\} \]
because \( x + y \) is an integer, and since \( y \) is small, \( \{x\} + y = 1 \).

5. **Calculate \( x - x^2 + x \lfloor x \rfloor \)**:
We want to compute:
\[ x - x^2 + x \lfloor x \rfloor \]
Substitute \( \lfloor x \rfloor = x - \{x\} \):
\[ x - x^2 + x (x - \{x\}) = x - x^2 + x^2 - x\{x\} = x - x\{x\} \]

6. **Simplify**:
Since \( \{x\} = 1 - y \):
\[ x - x\{x\} = x - x(1 - y) = x - x + xy = xy \]

7. **Substituting \( y \)**:
We know \( xy = (5 + 2\sqrt{6})^n (5 - 2\sqrt{6})^n = ((5 + 2\sqrt{6})(5 - 2\sqrt{6}))^n = (25 - 24)^n = 1^n = 1 \).

8. **Final Result**:
Thus, we find:
\[ x - x^2 + x \lfloor x \rfloor = 1 \]

### Conclusion:
The value of \( x - x^2 + x \lfloor x \rfloor \) is \( \boxed{1} \).