Let the sum `sum_(n=1)^(9)1/(n(n+1)(n+2))` written in the rational form be `p/q` (where p and q are co-prime), then the value of `[(q-p)/10]` is (where [.] is the greatest integer function)

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{n=1}^{9} \frac{1}{n(n+1)(n+2)} \]

### Step 1: Simplify the fraction
We can start by simplifying the term \(\frac{1}{n(n+1)(n+2)}\) using partial fractions. We can express it as:

\[ \frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} \]

Multiplying through by the denominator \(n(n+1)(n+2)\) gives:

\[ 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1) \]

### Step 2: Solve for A, B, and C
Expanding the right-hand side:

\[ 1 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n) \]

Combining like terms, we have:

\[ 1 = (A + B + C)n^2 + (3A + 2B + C)n + 2A \]

Setting up the system of equations by comparing coefficients:

1. \(A + B + C = 0\)
2. \(3A + 2B + C = 0\)
3. \(2A = 1\)

From the third equation, we find \(A = \frac{1}{2}\). Substituting \(A\) into the first two equations:

1. \(\frac{1}{2} + B + C = 0 \implies B + C = -\frac{1}{2}\)
2. \(3(\frac{1}{2}) + 2B + C = 0 \implies \frac{3}{2} + 2B + C = 0\)

From the second equation, substituting \(C = -\frac{1}{2} - B\):

\[ \frac{3}{2} + 2B - \frac{1}{2} - B = 0 \implies B + 1 = 0 \implies B = -1 \]

Substituting \(B = -1\) back into \(B + C = -\frac{1}{2}\):

\[ -1 + C = -\frac{1}{2} \implies C = \frac{1}{2} \]

Thus, we have:

\[ A = \frac{1}{2}, \quad B = -1, \quad C = \frac{1}{2} \]

### Step 3: Rewrite the original sum
Now we can rewrite our original sum \(S\):

\[ S = \sum_{n=1}^{9} \left( \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} \right) \]

### Step 4: Split the sum
This can be split into three separate sums:

\[ S = \frac{1}{2} \sum_{n=1}^{9} \frac{1}{n} - \sum_{n=1}^{9} \frac{1}{n+1} + \frac{1}{2} \sum_{n=1}^{9} \frac{1}{n+2} \]

### Step 5: Calculate each sum
Calculating each of these sums:

1. \(\sum_{n=1}^{9} \frac{1}{n} = H_9\) (the 9th harmonic number)
2. \(\sum_{n=1}^{9} \frac{1}{n+1} = H_{10} - 1\)
3. \(\sum_{n=1}^{9} \frac{1}{n+2} = H_{11} - 1 - \frac{1}{2}\)

### Step 6: Combine and simplify
Combining these results, we find:

\[ S = \frac{1}{2} H_9 - (H_{10} - 1) + \frac{1}{2} (H_{11} - \frac{3}{2}) \]

### Step 7: Final calculations
Calculating \(H_9\), \(H_{10}\), and \(H_{11}\):

- \(H_9 = 2.828968\)
- \(H_{10} = 3.028968\)
- \(H_{11} = 3.228968\)

Substituting these values into our expression for \(S\) and simplifying gives us \(S = \frac{27}{110}\).

### Step 8: Find \(p\) and \(q\)
Here, \(p = 27\) and \(q = 110\). They are co-prime.

### Step 9: Calculate \(\left\lfloor \frac{q - p}{10} \right\rfloor\)
Calculating:

\[ q - p = 110 - 27 = 83 \]

Thus:

\[ \left\lfloor \frac{83}{10} \right\rfloor = \left\lfloor 8.3 \right\rfloor = 8 \]

### Final Answer
The final answer is:

\[ \boxed{8} \]