1.1.2. \( \frac{6^{x+1} .9^{x-1}}{2^{x+1} .27^{x}} \)

To simplify the expression \( \frac{6^{x+1} \cdot 9^{x-1}}{2^{x+1} \cdot 27^{x}} \), we start with the factorizations of each term: - \( 6 = 2 \cdot 3 \), so \( 6^{x+1} = (2 \cdot 3)^{x+1} = 2^{x+1} \cdot 3^{x+1} \). - \( 9 = 3^2 \), thus \( 9^{x-1} = (3^2)^{x-1} = 3^{2(x-1)} = 3^{2x-2} \). - \( 2^{x+1} \) remains as is. - \( 27 = 3^3 \), therefore \( 27^{x} = (3^3)^{x} = 3^{3x} \). Now substituting these into the expression, we get: \[ \frac{2^{x+1} \cdot 3^{x+1} \cdot 3^{2x-2}}{2^{x+1} \cdot 3^{3x}} \] This simplifies to: \[ \frac{2^{x+1} \cdot 3^{(x+1) + (2x-2)}}{2^{x+1} \cdot 3^{3x}} = \frac{2^{x+1} \cdot 3^{3x - 1}}{2^{x+1} \cdot 3^{3x}} \] The \( 2^{x+1} \) cancels out, leading to: \[ \frac{3^{3x-1}}{3^{3x}} = 3^{-1} = \frac{1}{3} \] Thus, the simplified form of the expression is: \[ \frac{1}{3} \]