If `f(x)= |{:(1,,x,,x+1),(2x,,x(x-1),,(x+1)x),(3x(x-1),,x(x-1)(x-2),,(x+1)x(x-1)):}|` then the value of f(500) `"_____"`

Explanation

To solve the problem, we need to evaluate the determinant given by the function \( f(x) \):

\[ f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{array} \right| \]

### Step 1: Factor out common terms from the rows

We can factor out common terms from the second and third rows:

- From the second row, we can factor out \( x \).
- From the third row, we can factor out \( x(x - 1) \).

This gives us:

\[ f(x) = x \cdot x(x - 1) \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2 & (x - 1) & (x + 1) \\ 3 & (x - 2) & (x + 1) \end{array} \right| \]

### Step 2: Simplify the determinant

Now we simplify the determinant:

\[ f(x) = x^2(x - 1) \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2 & (x - 1) & (x + 1) \\ 3 & (x - 2) & (x + 1) \end{array} \right| \]

Next, we can perform column operations to simplify the determinant. We can replace the third column \( C_3 \) with \( C_3 - C_2 \):

\[ f(x) = x^2(x - 1) \left| \begin{array}{ccc} 1 & x & 1 \\ 2 & (x - 1) & 2 \\ 3 & (x - 2) & 3 \end{array} \right| \]

### Step 3: Calculate the determinant

Now, we can calculate the determinant:

\[ \left| \begin{array}{ccc} 1 & x & 1 \\ 2 & (x - 1) & 2 \\ 3 & (x - 2) & 3 \end{array} \right| \]

Notice that the first column and the third column are linearly dependent (the third column is a multiple of the first). Therefore, the determinant is zero:

\[ \left| \begin{array}{ccc} 1 & x & 1 \\ 2 & (x - 1) & 2 \\ 3 & (x - 2) & 3 \end{array} \right| = 0 \]

### Step 4: Conclusion

Thus, we can conclude that:

\[ f(x) = x^2(x - 1) \cdot 0 = 0 \]

So, for \( f(500) \):

\[ f(500) = 0 \]

### Final Answer

The value of \( f(500) \) is:

\[ \boxed{0} \]