Simplity the -lollowing \( \frac{2^{x+2}-2^{x+1}}{2^{x}+2^{x+2}} \quad \frac{b) 2 \cdot 3^{x}+3^{x-2}}{5 \cdot 3^{x+1}-7 \cdot 3^{x-1}} \)

To simplify the first expression \( \frac{2^{x+2}-2^{x+1}}{2^{x}+2^{x+2}} \), we can factor out common terms: The numerator can be simplified as: \( 2^{x+2} - 2^{x+1} = 2^{x+1}(2 - 1) = 2^{x+1} \). The denominator can be simplified as: \( 2^{x} + 2^{x+2} = 2^{x}(1 + 2^{2}) = 2^{x}(1 + 4) = 5 \cdot 2^{x} \). So the entire expression simplifies to: \[ \frac{2^{x+1}}{5 \cdot 2^{x}} = \frac{2^{x+1}}{5 \cdot 2^{x}} = \frac{2}{5} \] Now for the second expression \( \frac{2 \cdot 3^{x}+3^{x-2}}{5 \cdot 3^{x+1}-7 \cdot 3^{x-1}} \): We can start by simplifying the numerator: \( 2 \cdot 3^{x} + 3^{x-2} = 3^{x-2}(2 \cdot 3^{2} + 1) = 3^{x-2}(18 + 1) = 19 \cdot 3^{x-2} \). Then simplify the denominator: \( 5 \cdot 3^{x+1} - 7 \cdot 3^{x-1} = 5 \cdot 3^{x+1} - 7/3 \cdot 3^{x} = 5 \cdot 3^{x} \cdot 3 - 7 \cdot 3^{x} = (15 - 7) \cdot 3^{x} = 8 \cdot 3^{x} \). Putting it together, we have: \[ \frac{19 \cdot 3^{x-2}}{8 \cdot 3^{x}} = \frac{19}{8} \cdot \frac{1}{3^{2}} = \frac{19}{72} \] So, 1. The first simplified expression is \( \frac{2}{5} \). 2. The second simplified expression is \( \frac{19}{72} \).