\( \begin{array}{ll}\text { (4) } \frac{5^{x+4}-5^{x+3}}{100.5^{x+1}} & \text { (5) } \frac{4^{x}+3 \cdot 2^{2 x+1}}{7 \cdot 2^{2 x+1}} \\ \begin{array}{lll}\text { (7) } \frac{8^{x} \cdot 2^{x}+2.16^{x+1}}{11.2^{x+1}} & \text { (8) } \frac{12^{x}+4^{x} \cdot 3^{x+1}}{2^{2 x+4} \cdot 3^{x}} & \text { (6) } \frac{\left(3^{x}\right)^{2}-9^{x-1}}{9^{x-1}} \\ 5.2^{x+1}-7 \cdot 3^{x-1}\end{array} \\ 37\end{array} \)

To solve the given expressions, we will simplify each one step by step. Let's denote the expressions as follows: 1. \( E_4 = \frac{5^{x+4}-5^{x+3}}{100 \cdot 5^{x+1}} \) 2. \( E_5 = \frac{4^{x}+3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} \) 3. \( E_6 = \frac{(3^{x})^{2}-9^{x-1}}{9^{x-1}} \) 4. \( E_7 = \frac{8^{x} \cdot 2^{x}+2 \cdot 16^{x+1}}{11 \cdot 2^{x+1}} \) 5. \( E_8 = \frac{12^{x}+4^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} \) Let's simplify each expression one by one. ### Simplifying \( E_4 \) \[ E_4 = \frac{5^{x+4} - 5^{x+3}}{100 \cdot 5^{x+1}} \] Factoring out \( 5^{x+3} \) from the numerator: \[ = \frac{5^{x+3}(5 - 1)}{100 \cdot 5^{x+1}} = \frac{4 \cdot 5^{x+3}}{100 \cdot 5^{x+1}} \] Now simplifying: \[ = \frac{4 \cdot 5^{2}}{100} = \frac{100}{100} = 1 \] ### Simplifying \( E_5 \) \[ E_5 = \frac{4^{x} + 3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} \] Rewriting \( 4^{x} \) as \( (2^2)^{x} = 2^{2x} \): \[ = \frac{2^{2x} + 3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} = \frac{2^{2x} + 6 \cdot 2^{2x}}{7 \cdot 2^{2x+1}} \] Factoring out \( 2^{2x} \): \[ = \frac{7 \cdot 2^{2x}}{7 \cdot 2^{2x+1}} = \frac{1}{2} \] ### Simplifying \( E_6 \) \[ E_6 = \frac{(3^{x})^{2} - 9^{x-1}}{9^{x-1}} \] Rewriting \( 9^{x-1} \) as \( (3^2)^{x-1} = 3^{2x-2} \): \[ = \frac{3^{2x} - 3^{2x-2}}{3^{2x-2}} = \frac{3^{2x} - \frac{1}{9} \cdot 3^{2x}}{3^{2x-2}} \] Factoring out \( 3^{2x} \): \[ = \frac{3^{2x}(1 - \frac{1}{9})}{3^{2x-2}} = \frac{3^{2x} \cdot \frac{8}{9}}{3^{2x-2}} = \frac{8 \cdot 3^{2}}{9} = \frac{72}{9} = 8 \] ### Simplifying \( E_7 \) \[ E_7 = \frac{8^{x} \cdot 2^{x} + 2 \cdot 16^{x+1}}{11 \cdot 2^{x+1}} \] Rewriting \( 8^{x} \) and \( 16^{x+1} \): \[ = \frac{(2^3)^{x} \cdot 2^{x} + 2 \cdot (2^4)^{x+1}}{11 \cdot 2^{x+1}} = \frac{2^{4x} + 2 \cdot 2^{4x + 4}}{11 \cdot 2^{x+1}} \] Factoring out \( 2^{4x} \): \[ = \frac{2^{4x}(1 + 2^{5})}{11 \cdot 2^{x+1}} = \frac{2^{4x} \cdot 33}{11 \cdot 2^{x+1}} = \frac{3 \cdot 2^{4x}}{2^{x+1}} = 3 \cdot 2^{3x-1} \] ### Simplifying \( E_8 \) \[ E_8 = \frac{12^{x} + 4^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} \] Rewriting \( 12^{x} \) and \( 4^{x} \): \[ = \frac{(2^2 \cdot 3)^{x} + (2^2)^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} = \frac{2^{2x} \cdot 3^{x} + 2^{2x} \cdot 3^{x} \cdot 3}{2^{2x+4} \cdot 3^{x}} \] Factoring out \( 2^{2x} \cdot 3^{x} \): \[ = \frac{2^{2x} \cdot 3^{x}(1 + 3)}{2^{2x+4} \cdot 3^{x}} = \frac{4 \cdot 2^{2x}}{2^{2x+4}} = \frac{4}{16} = \frac{1}{4} \] ### Summary of Results 1. \( E_4 = 1 \) 2. \( E_5 = \frac{1}{2} \) 3. \( E_6 = 8 \) 4. \( E_7 = 3 \cdot 2^{3x-1} \) 5. \( E_8 = \frac{1}{4} \) Now, we can summarize the results or proceed with any further calculations if needed. If you have any