Solve:
`(|x-3|)/(x-3)gt0,x in R`

Explanation

To solve the inequality \(\frac{|x-3|}{x-3} > 0\), we will break it down step by step.

### Step 1: Understand the Absolute Value
The absolute value function \(|x-3|\) can be expressed in two cases:
1. When \(x - 3 \geq 0\) (i.e., \(x \geq 3\)), \(|x-3| = x - 3\).
2. When \(x - 3 < 0\) (i.e., \(x < 3\)), \(|x-3| = -(x - 3) = 3 - x\).

### Step 2: Analyze the Cases
We will analyze the inequality \(\frac{|x-3|}{x-3} > 0\) for both cases.

**Case 1:** \(x \geq 3\)

In this case, \(|x-3| = x - 3\). Therefore, we have:
\[ \frac{x-3}{x-3} > 0 \]
This simplifies to:
\[ 1 > 0 \]
This is true for all \(x > 3\). However, at \(x = 3\), the expression is undefined because the denominator becomes zero. Thus, we have:
\[ x > 3 \]

**Case 2:** \(x < 3\)

In this case, \(|x-3| = 3 - x\). Therefore, we have:
\[ \frac{3-x}{x-3} > 0 \]
This can be rewritten as:
\[ \frac{-(x-3)}{x-3} > 0 \]
This simplifies to:
\[ -1 > 0 \]
This is never true. Thus, there are no solutions for \(x < 3\).

### Step 3: Combine the Results
From both cases, we conclude that the only solution comes from Case 1:
\[ x > 3 \]

### Final Solution
The solution to the inequality \(\frac{|x-3|}{x-3} > 0\) is:
\[ x \in (3, \infty) \]