Simplify the following \( \begin{array}{lll}\text { (a) } \frac{10^{x} \cdot 25^{x+1}}{5^{5} \cdot 50^{x-1}} & \text { (b) } \frac{6^{n+2} \times 10^{n-2}}{4^{n} \times 15^{n-2}} & \text { (c) } \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} \\ \text { (d) } \frac{8^{n} .6^{-3} 9^{1-n}}{16^{n-1} \cdot 3^{-\pi}} & \text { (c) } \frac{2^{x+2}-2^{x+3}}{2^{x+1}-2^{x+2}} & \text { (f) } \frac{9^{x}+3^{2+1}}{18^{x} \cdot 2^{1-x}} \\ \text { (g) } \frac{3.2^{x}-2^{x-1}}{2^{1}+2^{x-2}} & \text { (h) } \frac{4^{x}+2^{2 x-1}}{2^{2 x-1}} & \text { (i) } \frac{2.3^{x+2}+3^{x-3}}{5.3^{x-2}}\end{array} \)

Let's simplify each expression step by step: (a) \[ \frac{10^{x} \cdot 25^{x+1}}{5^{5} \cdot 50^{x-1}} = \frac{(2 \cdot 5)^{x} \cdot (5^2)^{x+1}}{5^{5} \cdot (2 \cdot 5^2)^{x-1}} = \frac{2^{x} \cdot 5^{x} \cdot 5^{2x+2}}{5^{5} \cdot 2^{x-1} \cdot 5^{2(x-1)}} = \frac{2^{x} \cdot 5^{3x+2}}{2^{x-1} \cdot 5^{5}} = 2^{2} \cdot 5^{3x - 3} = 4 \cdot 5^{3x - 3} \] (b) \[ \frac{6^{n+2} \times 10^{n-2}}{4^{n} \times 15^{n-2}} = \frac{(2 \cdot 3)^{n+2} \cdot (2 \cdot 5)^{n-2}}{(2^2)^{n} \cdot (3 \cdot 5)^{n-2}} = \frac{2^{2} \cdot 3^{n+2} \cdot 2^{n-2} \cdot 5^{n-2}}{2^{2n} \cdot 3^{n-2} \cdot 5^{n-2}} = \frac{2^{n} \cdot 3^{n+4}}{2^{2n} \cdot 3^{n-2}} = \frac{3^{n+4}}{2^{n} \cdot 5^{n-2}} \] (c) \[ \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} = \frac{(2 \cdot 3)^{x} \cdot (3^2)^{x+1} \cdot 2}{(3^3)^{x+1}\cdot 2^{x-1}} = \frac{2^{x+1} \cdot 3^{3x + 2}}{2^{x-1} \cdot 3^{3x + 3}} = 2^2 \cdot 3^{-1} = \frac{4}{3} \] (d) \[ \frac{8^{n} \cdot 6^{-3} \cdot 9^{1-n}}{16^{n-1} \cdot 3^{-\pi}} = \frac{(2^3)^{n} \cdot (2 \cdot 3)^{-3} \cdot (3^2)^{1-n}}{(2^4)^{n-1} \cdot 3^{-\pi}} = \frac{2^{3n} \cdot 3^{-3} \cdot 3^{2 - 2n}}{2^{4n-4} \cdot 3^{\pi}} = \frac{2^{3n - 4n + 4} \cdot 3^{2 - 3 - \pi}}{1} = 2^{-n + 4} \cdot 3^{-1 - \pi} \] (e) \[ \frac{2^{x+2} - 2^{x+3}}{2^{x+1} - 2^{x+2}} = \frac{2^{x+2}(1 - 2)}{2^{x+1}(1 - 2)} = \frac{2^{x+2}(-1)}{2^{x+1}(-1)} = \frac{2^{x+2}}{2^{x+1}} = 2 \] (f) \[ \frac{9^{x}+3^{2+1}}{18^{x} \cdot 2^{1-x}} = \frac{(3^2)^{x}+3^{3}}{(2 \cdot 3^2)^{x} \cdot 2^{1-x}} = \frac{3^{2x}+3^{3}}{2^{x} \cdot