a) \( \frac{2^{2 x}-2}{2^{x}-1} \)

To simplify the expression \( \frac{2^{2 x}-2}{2^{x}-1} \), we can start by recognizing that \( 2^{2x} = (2^x)^2 \). Thus, we can rewrite the numerator \( 2^{2x} - 2 \) as \( (2^x)^2 - 2 \). Next, we can factor the expression as follows: 1. Notice that \( 2^{2x} - 2 = (2^x)^2 - (2^1) \), which has the form of a difference of squares. We can use the difference of squares formula: \( a^2 - b^2 = (a-b)(a+b) \). 2. Let \( a = 2^x \) and \( b = \sqrt{2} \). So, we get \( (2^x - \sqrt{2})(2^x + \sqrt{2}) \) for the numerator. Now our expression looks like this: \[ \frac{(2^x - \sqrt{2})(2^x + \sqrt{2})}{2^x - 1} \] For the final simplification, if \( 2^x - 1 \) is not equal to zero, we can leave it in this factored form. If \( 2^x - 1 = 0 \) (which happens when \( x = 0 \)), then this is the point of discontinuity. This gives us the simplified expression and an understanding of the behavior at specific values of \( x \). Happy calculating!