If `alpha` is a the absolute maximum value of the expression `(3x^(2)+2x-1)/(x^(2)+x+1)AA x in R`, then `[alpha]` is ______________. (where `[.]` denotes the greatest integer function)

Answer

To find the absolute maximum value of the expression \(\frac{3x^2 + 2x - 1}{x^2 + x + 1}\) for \(x \in \mathbb{R}\), we will follow these steps:

### Step 1: Set the expression equal to \(k\)
Let:
\[ y = \frac{3x^2 + 2x - 1}{x^2 + x + 1} = k \]
This implies:
\[ 3x^2 + 2x - 1 = k(x^2 + x + 1) \]

### Step 2: Rearrange the equation
Rearranging gives:
\[ 3x^2 + 2x - 1 - kx^2 - kx - k = 0 \]
This simplifies to:
\[ (3 - k)x^2 + (2 - k)x + (-1 - k) = 0 \]

### Step 3: Find the condition for real roots
For \(x\) to be real, the discriminant of this quadratic must be non-negative:
\[ D = (2 - k)^2 - 4(3 - k)(-1 - k) \geq 0 \]
Calculating the discriminant:
\[ D = (2 - k)^2 - 4(3 - k)(-1 - k) \]
Expanding this:
\[ D = (2 - k)^2 + 4(3 - k)(1 + k) \]
\[ = (2 - k)^2 + 4(3 + 3k - k - k^2) \]
\[ = (2 - k)^2 + 4(3 + 2k - k^2) \]

### Step 4: Set the discriminant greater than or equal to zero
Now we need to solve:
\[ (2 - k)^2 + 4(3 + 2k - k^2) \geq 0 \]

### Step 5: Simplify the inequality
Expanding and simplifying:
\[ (2 - k)^2 + 12 + 8k - 4k^2 \geq 0 \]
This leads to:
\[ -k^2 + 6k + 16 \geq 0 \]
Rearranging gives:
\[ k^2 - 6k - 16 \leq 0 \]

### Step 6: Find the roots of the quadratic
Using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2} \]
This gives:
\[ k = 8 \quad \text{and} \quad k = -2 \]

### Step 7: Determine the interval for \(k\)
The inequality \(k^2 - 6k - 16 \leq 0\) holds between the roots:
\[ -2 \leq k \leq 8 \]

### Step 8: Find the maximum value of \(k\)
The maximum value of \(k\) is \(8\).

### Step 9: Apply the greatest integer function
Thus, the absolute maximum value \(\alpha = 8\). The greatest integer function \([ \alpha ]\) is:
\[ [ \alpha ] = [8] = 8 \]

### Final Answer
The answer is:
\[ \boxed{8} \]