Solve : `(1)/(x) lt (2)/(x-2) `

Answer

To solve the inequality \( \frac{1}{x} < \frac{2}{x-2} \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \frac{1}{x} < \frac{2}{x-2} \]

### Step 2: Bring All Terms to One Side
To combine the fractions, we can rewrite the inequality as:
\[ \frac{1}{x} - \frac{2}{x-2} < 0 \]

### Step 3: Find a Common Denominator
The common denominator for the fractions is \( x(x-2) \). Thus, we rewrite the left side:
\[ \frac{(x-2) - 2x}{x(x-2)} < 0 \]

### Step 4: Simplify the Numerator
Now, simplify the numerator:
\[ \frac{x - 2 - 2x}{x(x-2)} < 0 \implies \frac{-x - 2}{x(x-2)} < 0 \]
This can be rewritten as:
\[ \frac{-(x + 2)}{x(x-2)} < 0 \]

### Step 5: Remove the Negative Sign
Multiplying both sides of the inequality by -1 (which reverses the inequality sign):
\[ \frac{x + 2}{x(x-2)} > 0 \]

### Step 6: Identify Critical Points
Now we need to find the critical points where the expression is zero or undefined:
1. \( x + 2 = 0 \) gives \( x = -2 \)
2. \( x = 0 \) (denominator)
3. \( x - 2 = 0 \) gives \( x = 2 \)

### Step 7: Test Intervals
We will test the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 0) \), \( (0, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ \frac{-3 + 2}{-3(-3-2)} = \frac{-1}{15} < 0 \quad \text{(not valid)} \]

2. **Interval \( (-2, 0) \)**: Choose \( x = -1 \)
\[ \frac{-1 + 2}{-1(-1-2)} = \frac{1}{3} > 0 \quad \text{(valid)} \]

3. **Interval \( (0, 2) \)**: Choose \( x = 1 \)
\[ \frac{1 + 2}{1(1-2)} = \frac{3}{-1} < 0 \quad \text{(not valid)} \]

4. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ \frac{3 + 2}{3(3-2)} = \frac{5}{3} > 0 \quad \text{(valid)} \]

### Step 8: Combine Valid Intervals
The valid intervals where the inequality holds are:
\[ (-2, 0) \quad \text{and} \quad (2, \infty) \]

### Final Answer
Thus, the solution to the inequality is:
\[ x \in (-2, 0) \cup (2, \infty) \]