(11.) Simplify ((a)) \( \left.\left\{\left\{\frac{x^{\frac{a}{a-b}}}{\left\{\frac{a}{x^{a+b}}\right.}\right\}\right\}\left\{\frac{x^{\frac{b}{b-a}}}{x^{\frac{b}{b+a}}}\right\}\right\}^{a+b} \)

To simplify the given expression \( \left\{\left\{\frac{x^{\frac{a}{a-b}}}{\left\{\frac{a}{x^{a+b}}\right.}\right\}\right\}\left\{\frac{x^{\frac{b}{b-a}}}{x^{\frac{b}{b+a}}}\right\}\right\}^{a+b} \), we can break it down step by step. 1. **Simplifying each part**: - For the first part, \( \frac{x^{\frac{a}{a-b}}}{\frac{a}{x^{a+b}}} = \frac{x^{\frac{a}{a-b}} \cdot x^{a+b}}{a} = \frac{x^{\frac{a}{a-b} + a + b}}{a} \). - This simplifies to \( \frac{x^{\frac{a + (a+b)(a-b)}{a-b}}}{a} \). 2. **Second part**: - For the second part, \( \frac{x^{\frac{b}{b-a}}}{x^{\frac{b}{b+a}}} = x^{\frac{b}{b-a} - \frac{b}{b+a}} = x^{b \left(\frac{1}{b-a} - \frac{1}{b+a}\right)} \). - Let’s simplify this expression: \[ \frac{1}{b-a} - \frac{1}{b+a} = \frac{(b+a) - (b-a)}{(b-a)(b+a)} = \frac{2a}{(b-a)(b+a)} \] Thus, the second part becomes \( x^{\frac{2ab}{(b-a)(b+a)}} \). 3. **Combining parts**: Combining the two simplified results, \[ \left(\frac{x^{\frac{a + (a+b)(a-b)}{a-b}}{a} \cdot x^{\frac{2ab}{(b-a)(b+a)}}}\right)^{a+b} \] which gives, \[ \left(\frac{x^{\left(\frac{a + (a+b)(a-b) + \frac{2a}{(b-a)(b+a)}}{a-b}\right)}{a}}\right)^{a+b} \] 4. **Final form**: The entire expression raised to the power \( a + b \) ultimately leads to: \[ = \left(\frac{x^{\frac{N}{D}}}{a}\right)^{a+b} \] where \( N = a + (a+b)(a-b) + 2a \) and \( D = a - b \). This entire process simplifies to terms that can be further worked upon, but provides the framework for evaluating the simplified expressions. In final simplified form, it can equate to: \[ \frac{x^{P}}{a^{a+b}} \text{ where P is the combined exponent derived from above.} \]