\( \begin{array}{ll}\frac{2^{2 x}-2^{x}}{2^{x}-1} & \text { (b) } \frac{2^{2 x}-1}{2^{x}+1} \\ \frac{16-4^{x}}{2^{x}-4} & \text { (c) } \frac{9^{x}-9}{3^{x}-3}\end{array} \)

Let's simplify each of the given expressions step by step. --- ### (a) \(\displaystyle \frac{2^{2x} - 2^{x}}{2^{x} - 1}\) **Step 1:** Factor the numerator. \[ 2^{2x} - 2^{x} = 2^{x} \cdot 2^{x} - 2^{x} = 2^{x}(2^{x} - 1) \] **Step 2:** Substitute back into the fraction. \[ \frac{2^{x}(2^{x} - 1)}{2^{x} - 1} \] **Step 3:** Cancel out the common term \((2^{x} - 1)\). \[ = 2^{x} \quad \text{(provided \(2^{x} \neq 1\), i.e., \(x \neq 0\))} \] --- ### (b) \(\displaystyle \frac{2^{2x} - 1}{2^{x} + 1}\) **Step 1:** Recognize the numerator as a difference of squares. \[ 2^{2x} - 1 = (2^{x})^2 - 1^2 = (2^{x} - 1)(2^{x} + 1) \] **Step 2:** Substitute back into the fraction. \[ \frac{(2^{x} - 1)(2^{x} + 1)}{2^{x} + 1} \] **Step 3:** Cancel out the common term \((2^{x} + 1)\). \[ = 2^{x} - 1 \] --- ### (c) \(\displaystyle \frac{9^{x} - 9}{3^{x} - 3}\) **Step 1:** Rewrite the terms with a common base. \[ 9^{x} = (3^2)^{x} = 3^{2x} \quad \text{and} \quad 9 = 3^2 \] **Step 2:** Factor the numerator as a difference of squares. \[ 3^{2x} - 3^2 = (3^{x})^2 - 3^2 = (3^{x} - 3)(3^{x} + 3) \] **Step 3:** Substitute back into the fraction. \[ \frac{(3^{x} - 3)(3^{x} + 3)}{3^{x} - 3} \] **Step 4:** Cancel out the common term \((3^{x} - 3)\). \[ = 3^{x} + 3 \quad \text{(provided \(3^{x} \neq 3\), i.e., \(x \neq 1\))} \] --- ### Additional Expression: \(\displaystyle \frac{16 - 4^{x}}{2^{x} - 4}\) **Step 1:** Rewrite the terms with a common base. \[ 4^{x} = (2^2)^{x} = 2^{2x} \quad \text{and} \quad 16 = 2^4 \] **Step 2:** Factor the numerator. \[ 16 - 4^{x} = 2^4 - 2^{2x} = -(2^{2x} - 2^4) = -\left(2^{x} - 4\right)\left(2^{x} + 4\right) \] **Step 3:** Substitute back into the fraction. \[ \frac{-(2^{x} - 4)(2^{x} + 4)}{2^{x} - 4} \] **Step 4:** Cancel out the common term \((2^{x} - 4)\). \[ = -\left(2^{x} + 4\right) \] --- ### Summary of Simplified Expressions 1. **(a)** \(\displaystyle \frac{2^{2x} - 2^{x}}{2^{x} - 1} = 2^{x}\) \quad (for \(x \neq 0\)) 2. **(b)** \(\displaystyle \frac{2^{2x} - 1}{2^{x} + 1} = 2^{x} - 1\) 3. **(c)** \(\displaystyle \frac{9^{x} - 9}{3^{x} - 3} = 3^{x} + 3\) 4. **Additional:** \(\displaystyle \frac{16 - 4^{x}}{2^{x} - 4} = -\left(2^{x} + 4\right)\) --- These simplifications assume that the denominators are not equal to zero, which imposes the following restrictions on \(x\): - For **(a)**: \(2^{x} \neq 1 \Rightarrow x \neq 0\) - For **(c)**: \(3^{x} \neq 3 \Rightarrow x \neq 1\) - For the **additional expression**: \(2^{x} \neq 4 \Rightarrow x \neq 2\) If you have any further questions or need additional assistance, feel free to ask!