Solve : `(|x-3|)/(x^(2)-5x+6) gt 2`

Answer

To solve the inequality \(\frac{|x-3|}{x^2 - 5x + 6} > 2\), we will follow these steps:

### Step 1: Factor the denominator
First, we need to factor the quadratic expression in the denominator:
\[ x^2 - 5x + 6 = (x-2)(x-3) \]
So, we can rewrite the inequality as:
\[ \frac{|x-3|}{(x-2)(x-3)} > 2 \]

### Step 2: Consider cases based on the absolute value
The absolute value \(|x-3|\) leads us to consider two cases:

**Case 1:** \(x \geq 3\)
Here, \(|x-3| = x-3\). The inequality becomes:
\[ \frac{x-3}{(x-2)(x-3)} > 2 \]
Since \(x \neq 3\) (to avoid division by zero), we can simplify:
\[ \frac{1}{x-2} > 2 \]
This leads to:
\[ 1 > 2(x-2) \implies 1 > 2x - 4 \implies 5 > 2x \implies x < \frac{5}{2} \]
However, since we assumed \(x \geq 3\), there are no valid solutions in this case.

**Case 2:** \(x < 3\)
Here, \(|x-3| = 3-x\). The inequality becomes:
\[ \frac{3-x}{(x-2)(3-x)} > 2 \]
Again, we can simplify:
\[ \frac{1}{x-2} > 2 \]
This leads to:
\[ 1 > 2(x-2) \implies 1 > 2x - 4 \implies 5 > 2x \implies x < \frac{5}{2} \]
Now, we need to consider the restriction from the denominator:
- \(x \neq 2\) (to avoid division by zero)
- \(x < 3\)

### Step 3: Combine the results
From Case 2, we found \(x < \frac{5}{2}\). Thus, we need to combine this with the restriction \(x < 3\) and \(x \neq 2\). Therefore, the valid solution set is:
\[ \frac{3}{2} < x < 2 \]

### Final Solution
Thus, the solution to the inequality \(\frac{|x-3|}{x^2 - 5x + 6} > 2\) is:
\[ \boxed{\left(\frac{3}{2}, 2\right)} \]

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