if `(y^(2)-y+1)[x-2]lty^(2)+y+1,AAyepsilonR` then `([.]` denotes G.I.F) number of positive integer value of x is

Explanation

To solve the inequality \((y^{2} - y + 1)[\text{gif}(x - 2)] < y^{2} + y + 1\) where \(y \in \mathbb{R}\), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ (y^{2} - y + 1)[\text{gif}(x - 2)] < y^{2} + y + 1 \]
This can be rearranged to isolate the term involving \(\text{gif}(x - 2)\):
\[ \text{gif}(x - 2) < \frac{y^{2} + y + 1}{y^{2} - y + 1} \]

### Step 2: Analyze the right-hand side
Let \(k = \frac{y^{2} + y + 1}{y^{2} - y + 1}\). We need to find the range of \(k\) for \(y \in \mathbb{R}\).

### Step 3: Determine the bounds of \(k\)
To find the minimum and maximum values of \(k\), we can analyze the expression:
\[ k = 1 + \frac{2y}{y^{2} - y + 1} \]
We can find the critical points by differentiating \(k\) with respect to \(y\) and setting the derivative to zero. However, we can also evaluate the limits as \(y\) approaches certain values.

1. As \(y \to \infty\), \(k \to 1 + 2 = 3\).
2. As \(y \to -\infty\), \(k \to 1 - 2 = -1\) (but we need to check the minimum value).

By evaluating \(k\) at specific points, we can find that \(k\) is bounded between \(\frac{1}{3}\) and \(3\).

### Step 4: Set the bounds for \(\text{gif}(x - 2)\)
From the inequality \(\text{gif}(x - 2) < k\), we know:
\[ \text{gif}(x - 2) < \frac{1}{3} \quad \text{and} \quad \text{gif}(x - 2) < 3 \]
Since \(\text{gif}(x - 2)\) can take integer values, we focus on the inequality \(\text{gif}(x - 2) < \frac{1}{3}\). The only integer value that satisfies this is:
\[ \text{gif}(x - 2) \leq 0 \]

### Step 5: Determine possible values for \(x\)
The greatest integer function \(\text{gif}(x - 2)\) can take values \(0, -1, -2, -3, \ldots\).

1. If \(\text{gif}(x - 2) = 0\), then \(x - 2 < 1\) which gives \(x < 3\).
2. If \(\text{gif}(x - 2) = -1\), then \(x - 2 < 0\) which gives \(x < 2\).
3. If \(\text{gif}(x - 2) = -2\), then \(x - 2 < -1\) which gives \(x < 1\).

### Step 6: Find positive integer values of \(x\)
The positive integer values of \(x\) that satisfy \(x < 3\) are:
- \(x = 1\)
- \(x = 2\)

Thus, the total number of positive integer values of \(x\) that satisfy the original inequality is:
\[ \boxed{2} \]