Simplify the following: \( \begin{array}{lll}\text { (a) } \frac{2^{2 x}-2^{x}}{2^{x}-1} & \text { (b) } \frac{2^{2 x}-1}{2^{x}+1} & \text { (c) } \frac{9^{x}-9}{3^{x}-3} \\ \text { (d) } \frac{16-4^{x}}{2^{x}-4} & \end{array} \)

Let's simplify each expression step by step. (a) \( \frac{2^{2x} - 2^{x}}{2^{x} - 1} \) We can factor the numerator: \[ 2^{2x} - 2^{x} = 2^{x}(2^{x} - 1) \] Now substituting this back into the expression gives: \[ \frac{2^{x}(2^{x} - 1)}{2^{x} - 1} \] Assuming \( 2^{x} \neq 1 \) (which means \( x \neq 0 \)), we can cancel \( (2^{x} - 1) \): \[ = 2^{x} \] (b) \( \frac{2^{2x} - 1}{2^{x} + 1} \) The numerator can be factored using the difference of squares: \[ 2^{2x} - 1 = (2^{x} - 1)(2^{x} + 1) \] Thus, substituting this into the fraction gives: \[ \frac{(2^{x} - 1)(2^{x} + 1)}{2^{x} + 1} \] Assuming \( 2^{x} \neq -1 \) (valid since \( 2^{x} \) is always positive), we can cancel \( (2^{x} + 1) \): \[ = 2^{x} - 1 \] (c) \( \frac{9^{x} - 9}{3^{x} - 3} \) We can express \( 9^{x} \) as \( (3^{2})^{x} = (3^{x})^{2} \): \[ \frac{(3^{x})^{2} - 3^{2}}{3^{x} - 3} \] The numerator now factors as a difference of squares: \[ = \frac{(3^{x} - 3)(3^{x} + 3)}{3^{x} - 3} \] Assuming \( 3^{x} \neq 3 \), we can cancel \( (3^{x} - 3) \): \[ = 3^{x} + 3 \] (d) \( \frac{16 - 4^{x}}{2^{x} - 4} \) We can rewrite \( 4^{x} \) as \( (2^{2})^{x} = (2^{x})^{2} \), thus: \[ = \frac{16 - (2^{x})^{2}}{2^{x} - 4} \] The numerator is a difference of squares: \[ = \frac{(4 - 2^{x})(4 + 2^{x})}{2^{x} - 4} \] Noticing that \( 2^{x} - 4 = -(4 - 2^{x}) \), we have: \[ = -\frac{(4 - 2^{x})(4 + 2^{x})}{(4 - 2^{x})} = -(4 + 2^{x}) \] So the simplified forms of the expressions are: (a) \( 2^{x} \) (b) \( 2^{x} - 1 \) (c) \( 3^{x} + 3 \) (d) \( -(4 + 2^{x}) \)