Simplify a, \( \frac{3.2^{2 x}-2^{x-4}}{2^{x}+2^{x+2}} \)

To simplify the expression \( \frac{3.2^{2 x}-2^{x-4}}{2^{x}+2^{x+2}} \), we first note that \( 3.2^{2x} \) can be rewritten using properties of exponents. We know that \( 3.2^{2x} = 3 \cdot (2^{x})^2 = 3 \cdot 2^{2x} \). For the term \( 2^{x-4} \), it can be rewritten as \( \frac{2^x}{16} \). Now, let's rewrite the denominator: \[ 2^{x} + 2^{x+2} = 2^{x} + 4 \cdot 2^{x} = 5 \cdot 2^{x}. \] Putting this all together, the expression becomes: \[ \frac{3 \cdot 2^{2x} - \frac{2^x}{16}}{5 \cdot 2^x}. \] To further simplify the numerator: \[ 3 \cdot 2^{2x} - \frac{1}{16} \cdot 2^x. \] To combine the terms, we'll factor out \( 2^x \) from the numerator: \[ 2^x \left(3 \cdot 2^x - \frac{1}{16}\right). \] Now substituting this back into our original fraction, we have: \[ \frac{2^x (3 \cdot 2^x - \frac{1}{16})}{5 \cdot 2^x}. \] The \( 2^x \) in the numerator and denominator cancels out (assuming \( x \neq 0 \)): \[ \frac{3 \cdot 2^x - \frac{1}{16}}{5}. \] Thus, the simplified expression is: \[ \frac{3 \cdot 2^x - \frac{1}{16}}{5}. \]