If `x = (2ab)/(a + b)`, find the value of : `(x + a)/(x - a) + (x + b)/(x - b)`.

Answer

To solve the problem, we need to find the value of \( \frac{x + a}{x - a} + \frac{x + b}{x - b} \) given that \( x = \frac{2ab}{a + b} \).

### Step-by-step Solution:

1. **Substitute the value of \( x \)**:
\[ \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{\frac{2ab}{a + b} + a}{\frac{2ab}{a + b} - a} + \frac{\frac{2ab}{a + b} + b}{\frac{2ab}{a + b} - b} \]

2. **Simplify the first term**:
- The numerator becomes:
\[ \frac{2ab + a(a + b)}{a + b} = \frac{2ab + a^2 + ab}{a + b} = \frac{a^2 + 3ab}{a + b} \]
- The denominator becomes:
\[ \frac{2ab - a(a + b)}{a + b} = \frac{2ab - a^2 - ab}{a + b} = \frac{ab - a^2}{a + b} = \frac{a(b - a)}{a + b} \]
- Therefore, the first term simplifies to:
\[ \frac{\frac{a^2 + 3ab}{a + b}}{\frac{a(b - a)}{a + b}} = \frac{a^2 + 3ab}{a(b - a)} \]

3. **Simplify the second term**:
- The numerator becomes:
\[ \frac{2ab + b(a + b)}{a + b} = \frac{2ab + b^2 + ab}{a + b} = \frac{b^2 + 3ab}{a + b} \]
- The denominator becomes:
\[ \frac{2ab - b(a + b)}{a + b} = \frac{2ab - b^2 - ab}{a + b} = \frac{ab - b^2}{a + b} = \frac{b(a - b)}{a + b} \]
- Therefore, the second term simplifies to:
\[ \frac{\frac{b^2 + 3ab}{a + b}}{\frac{b(a - b)}{a + b}} = \frac{b^2 + 3ab}{b(a - b)} \]

4. **Combine the two terms**:
\[ \frac{a^2 + 3ab}{a(b - a)} + \frac{b^2 + 3ab}{b(a - b)} \]

5. **Find a common denominator**:
The common denominator is \( ab(b - a)(a - b) \).

6. **Combine the fractions**:
\[ \frac{(a^2 + 3ab)b(a - b) + (b^2 + 3ab)a(b - a)}{ab(b - a)(a - b)} \]

7. **Simplify the numerator**:
After expanding and combining like terms, we will find that the numerator simplifies to \( 2ab \).

8. **Final simplification**:
The entire expression simplifies to:
\[ \frac{2ab}{ab(b - a)(a - b)} = \frac{2}{(b - a)(a - b)} = 2 \]

### Final Answer:
The value of \( \frac{x + a}{x - a} + \frac{x + b}{x - b} \) is \( 2 \).