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The equation provided asks to solve for \( x \), and the equation is:

\[
(a + b) \left[ \frac{a + bx}{a + b} - \frac{a - bx}{a - b} \right] = \frac{abx}{a - b}
\]

Let's solve for \( x \).

### Step-by-Step Solution:

1. **Simplify the expression within the brackets**:
   \[
   \frac{a + bx}{a + b} - \frac{a - bx}{a - b}
   \]
   To subtract these fractions, find a common denominator:
   \[
   \text{Common denominator} = (a + b)(a - b)
   \]

2. **Combine the two fractions**:
   \[
   \frac{(a + bx)(a - b) - (a - bx)(a + b)}{(a + b)(a - b)}
   \]
   Now expand both terms in the numerator:
   \[
   (a + bx)(a - b) = a(a - b) + bx(a - b) = a^2 - ab + bxa - bxb
   \]
   \[
   (a - bx)(a + b) = a(a + b) - bx(a + b) = a^2 + ab - bxa - bxb
   \]
   Now subtract the two:
   \[
   (a^2 - ab + bxa - bxb) - (a^2 + ab - bxa - bxb)
   \]
   Simplify the expression:
   \[
   = -2ab + 2bxa
   \]
   Thus, the new expression becomes:
   \[
   \frac{-2ab + 2bxa}{(a + b)(a - b)}
   \]

3. **Substitute back into the original equation**:
   \[
   (a + b) \left( \frac{-2ab + 2bxa}{(a + b)(a - b)} \right) = \frac{abx}{a - b}
   \]

4. **Cancel out \( (a + b) \) on both sides**:
   \[
   \frac{-2ab + 2bxa}{a - b} = \frac{abx}{a - b}
   \]

5. **Multiply both sides by \( a - b \) to eliminate the denominator**:
   \[
   -2ab + 2bxa = abx
   \]

6. **Solve for \( x \)**:
   Collect terms involving \( x \) on one side:
   \[
   2bxa - abx = 2ab
   \]
   Factor out \( x \) on the left-hand side:
   \[
   x(2ba - ab) = 2ab
   \]
   Simplify the expression:
   \[
   x(ab) = 2ab
   \]
   Divide both sides by \( ab \):
   \[
   x = 2
   \]

Thus, the solution is:

\[
x = 2
\]

Would you like more details or further clarification on any step?

### Related Questions:
1. What is the significance of finding a common denominator when subtracting fractions?
2. How do you expand binomials such as \( (a + bx)(a - b) \)?
3. Can this method be used to solve equations involving higher degrees of \( x \)?
4. What are the conditions for canceling terms in an equation?
5. Why does factoring help in solving for \( x \) in equations like this?

### Tip:
When solving for variables in fractions, always look for opportunities to factor and simplify terms early to make the algebra easier.

Despeja la variable 'x' en la siguiente ecuación: (a + b) [ (a + bx) / (a + b) - (a - bx) / (a - b) ] = (abx) / (a - b)

Learn how to solve a complex algebraic equation involving fractions. This problem involves key algebraic concepts such as simplification, factoring, and solving for a variable, x. Suitable for students in Grades 10-12.

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