\( \begin{array}{ll}\text { 1.1.3 } & \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} \\ \text { 1.1.4 } & \frac{25^{x+1} \cdot 6}{10^{x-1} \cdot 15^{x}}\end{array} \)

To simplify \( \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} \), notice that you can factor both the numerator and the denominator. The numerator becomes \( 2^{x}(1 + 2) = 2^{x} \cdot 3 \), and the denominator can be rewritten as \( 2^{x}(1 - 4) = 2^{x} \cdot (-3) \). Therefore, the expression simplifies to \( \frac{2^{x} \cdot 3}{2^{x} \cdot (-3)} = -1 \). For the expression \( \frac{25^{x+1} \cdot 6}{10^{x-1} \cdot 15^{x}} \), we can break down each term using their prime factorizations. \( 25^{x+1} = (5^2)^{x+1} = 5^{2x+2} \), \( 10^{x-1} = (2 \cdot 5)^{x-1} = 2^{x-1} \cdot 5^{x-1} \), and \( 15^{x} = (3 \cdot 5)^{x} = 3^{x} \cdot 5^{x} \). The expression then becomes \( \frac{5^{2x+2} \cdot 6}{2^{x-1} \cdot 5^{x-1} \cdot 3^{x} \cdot 5^{x}} = \frac{6 \cdot 5^{2x+2}}{2^{x-1} \cdot 5^{2x-1} \cdot 3^{x}} \), simplifying ultimately to a more manageable fraction based on your variable values.