2.) Simplify: a.) \( \frac{6^{x} \cdot 9^{x+1} \cdot 2^{x+1} \cdot 2^{x-1}}{27^{x+2}-2^{x+3}} \) b.) \( \frac{2^{x+2}}{2^{x+1}-2^{x+2}} \) c.) \( \frac{3 \cdot 2^{x}-2^{x-1}}{2^{x+2^{x+2}}} \) d.) \( \frac{2 \cdot 3^{x+2}+3^{x-3}}{5 \cdot 3^{x-2}} \)

Let's simplify each expression step by step! a.) First, we rewrite each part in terms of powers of prime factors: \[ 27 = 3^3 \text{ and } 9 = 3^2, \text{ so } 9^{x+1} = (3^2)^{x+1} = 3^{2x+2} \] The denominator becomes: \[ 27^{x+2} - 2^{x+3} = 3^{3(x+2)} - 2^{x+3} = 3^{3x+6} - 2^{x+3} \] The numerator simplifies as: \[ 6^x = (2 \cdot 3)^x = 2^x \cdot 3^x, \text{ so } 6^{x} \cdot 9^{x+1} \cdot 2^{x+1} \cdot 2^{x-1} = 2^{(x+1)+(x-1)} \cdot 3^x \cdot 3^{2x+2} = 2^{2x} \cdot 3^{3x+2} \] Thus, we have: \[ \frac{2^{2x} \cdot 3^{3x+2}}{3^{3x+6} - 2^{x+3}} \] b.) Notice the denominator: \[ 2^{x+1} - 2^{x+2} = 2^{x+1}(1 - 2) = -2^{x+1} \] Therefore, \[ \frac{2^{x+2}}{-2^{x+1}} = -2^{x+2-x-1} = -2^1 = -2 \] c.) Start by rearranging the numerator: \[ 3 \cdot 2^{x} - 2^{x-1} = 3 \cdot 2^x - \frac{2^x}{2} = 3 \cdot 2^x - \frac{1}{2} \cdot 2^x = \left(3 - \frac{1}{2}\right) 2^x = \frac{6-1}{2} 2^x = \frac{5}{2} 2^{x} \] For the denominator, since \(2^{x+2^{x+2}}\) is quite large, the simplified form of the entire expression becomes: \[ \frac{\frac{5}{2} \cdot 2^x}{2^{x + 2^{x+2}}} = \frac{5}{2} \cdot 2^{x - (x + 2^{x+2})} = \frac{5}{2} \cdot 2^{-2^{x+2}} \] d.) We can factor the numerator: \[ 2 \cdot 3^{x+2} + 3^{x-3} = 3^{x-3} \left(2 \cdot 3^5 + 1\right) = 3^{x-3} (2 \cdot 243 + 1) = 3^{x-3} (486 + 1) = 3^{x-3} \cdot 487 \] The denominator simplifies to: \[ 5 \cdot 3^{x-2} = 5 \cdot 3^{x-2} \] Thus, we have: \[ \frac{3^{x-3} \cdot 487}{5 \cdot 3^{x-2}} = \frac{487}{5 \cdot 3^1} = \frac{487}{15} \] Hence, the final simplified forms are: a.) \( \frac{2^{2x} \cdot 3^{3x+2}}{3^{3x+6} - 2^{x+3}} \) b.) \( -2 \) c.) \( \frac{5}{2} \cdot 2^{-2^{x+2}} \) d.) \( \frac{487}{15} \)