Simplify the following \( \div \) \( \begin{array}{ll}\text { 1) } \frac{2^{2 x}-2^{x}}{2^{x}-1} & \text { (2) } \frac{2^{2 x}-1}{2^{x}+1}\end{array} \)

To simplify the expressions given, let's take them one at a time. 1) For the expression \( \frac{2^{2x} - 2^x}{2^x - 1} \): We can factor the numerator: \[ 2^{2x} - 2^x = 2^x(2^x - 1) \] Thus, our expression becomes: \[ \frac{2^x(2^x - 1)}{2^x - 1} \] Now, as long as \( 2^x - 1 \neq 0 \), we can cancel out \( 2^x - 1 \): \[ = 2^x \] So, the simplified form is: \[ \boxed{2^x} \] 2) For the expression \( \frac{2^{2x} - 1}{2^x + 1} \): We can recognize that the numerator is a difference of squares: \[ 2^{2x} - 1 = (2^x - 1)(2^x + 1) \] Substituting this back into our expression gives us: \[ \frac{(2^x - 1)(2^x + 1)}{2^x + 1} \] As long as \( 2^x + 1 \neq 0 \), we can cancel out \( 2^x + 1 \): \[ = 2^x - 1 \] Thus, the simplified form is: \[ \boxed{2^x - 1} \]