The value of ` [ ( 2009! + 2006!)/(2008! + 2007!)] ` =
(` [ *] ` denotes greatest integer function.)

Options

A. 2009
B. 2008
C. 2007
D. 1

Answer

To solve the expression \(\left\lfloor \frac{2009! + 2006!}{2008! + 2007!} \right\rfloor\), we can simplify the fraction step by step.

### Step 1: Factor out the common terms
We start by factoring out \(2006!\) from the numerator and \(2007!\) from the denominator.

\[ \frac{2009! + 2006!}{2008! + 2007!} = \frac{2006!(2009 \times 2008 \times 2007 + 1)}{2007!(2008 + 1)} \]

### Step 2: Rewrite the denominator
We know that \(2007! = 2007 \times 2006!\), so we can rewrite the denominator:

\[ = \frac{2006!(2009 \times 2008 \times 2007 + 1)}{2007 \times 2006!(2008 + 1)} \]

### Step 3: Cancel out \(2006!\)
Now, we can cancel \(2006!\) from the numerator and denominator:

\[ = \frac{2009 \times 2008 \times 2007 + 1}{2007 \times (2008 + 1)} \]

### Step 4: Simplify the expression
This simplifies to:

\[ = \frac{2009 \times 2008 \times 2007 + 1}{2007 \times 2009} \]

### Step 5: Divide the numerator by the denominator
Now we can divide the numerator by the denominator:

\[ = \frac{2008 \times 2007}{2007} + \frac{1}{2007 \times 2009} \]

This simplifies to:

\[ = 2008 + \frac{1}{2007 \times 2009} \]

### Step 6: Evaluate the greatest integer function
Now we need to find the greatest integer function of the expression:

\[ \left\lfloor 2008 + \frac{1}{2007 \times 2009} \right\rfloor \]

Since \(\frac{1}{2007 \times 2009}\) is a very small positive number, the value of \(2008 + \frac{1}{2007 \times 2009}\) will be slightly greater than \(2008\) but less than \(2009\).

Thus, the greatest integer function will yield:

\[ \left\lfloor 2008 + \text{(a small positive number)} \right\rfloor = 2008 \]

### Final Answer
Therefore, the final answer is:

\[ \boxed{2008} \]