1) \( \frac{27^{3-2 x} \cdot 9^{x-1}}{81^{2 x}} \) 2) \( \frac{6 \cdot 5^{x+1}-2 \cdot 5^{x-2}}{5^{x+3}} \) 3) \( \frac{2^{2009}-2^{2012}}{2^{2010}} \)

To simplify these expressions, we can use the properties of exponents and the fact that we can represent different bases as powers of the same base where applicable. 1) \( \frac{27^{3-2 x} \cdot 9^{x-1}}{81^{2 x}} \) First, express 27, 9, and 81 in terms of base 3: \( 27 = 3^3 \), \( 9 = 3^2 \), \( 81 = 3^4 \). So we can rewrite the expression as: \[ \frac{(3^3)^{3-2x} \cdot (3^2)^{x-1}}{(3^4)^{2x}} = \frac{3^{9-6x} \cdot 3^{2x-2}}{3^{8x}} = 3^{(9-6x + 2x - 2 - 8x)} = 3^{-5x + 7}. \] 2) \( \frac{6 \cdot 5^{x+1}-2 \cdot 5^{x-2}}{5^{x+3}} \) Factor out \( 5^{x-2} \) from the numerator: \[ 6 \cdot 5^{x+1} - 2 \cdot 5^{x-2} = 5^{x-2} \cdot (6 \cdot 5^3 - 2) = 5^{x-2} \cdot (6 \cdot 125 - 2) = 5^{x-2} \cdot 748. \] Thus, \[ \frac{5^{x-2} \cdot 748}{5^{x+3}} = 748 \cdot \frac{1}{5^5} = \frac{748}{5^5}. \] 3) \( \frac{2^{2009}-2^{2012}}{2^{2010}} \) Factor \( 2^{2009} \) out of the numerator: \[ \frac{2^{2009} (1 - 2^3)}{2^{2010}} = \frac{2^{2009} \cdot (-7)}{2^{2010}} = -7 \frac{2^{2009}}{2^{2010}} = -\frac{7}{2}. \] So the simplified results are: 1) \( 3^{-5x + 7} \) 2) \( \frac{748}{5^5} \) 3) \( -\frac{7}{2} \)