QUESTION 1 1.1 Fully simplify the following expressions: 1.1.1 \( \frac{19.3^{x}-9.3^{x+2}}{18.3^{x-3}+39.3^{1+x}} \) \( 1.1 .2 \frac{\sqrt{48 x^{6}}-\sqrt[3]{125 x^{9}} \cdot 3^{\frac{1}{2}}}{\sqrt{75 x^{6}}}=\frac{124}{2} \times ? \)

To simplify the first expression \( \frac{19.3^{x}-9.3^{x+2}}{18.3^{x-3}+39.3^{1+x}} \), we can factor out common terms: The numerator simplifies as follows: \[ 19.3^x - 9.3^{x+2} = 19.3^x - 9.3^x \cdot 3^2 = 3^x (19 - 81) = -62 \cdot 3^x \] The denominator simplifies like this: \[ 18.3^{x-3} + 39.3^{1+x} = 18.3^{x-3} + 39 \cdot 3^x = 3^{x-3}(18 + 39 \cdot 3^4) \rightarrow 3^{x-3}(18 + 39 \cdot 81) = 3^{x-3}(18 + 3159) \] Thus, \[ = 3^{x-3} \times 3177 \] So the entire expression becomes: \[ \frac{-62 \cdot 3^x}{3^{x-3} \cdot 3177} = \frac{-62 \cdot 3^3}{3177} = \frac{-62 \cdot 27}{3177} \] Which simplifies to about \(\approx -\frac{1674}{3177} \). For the second expression, \( \frac{\sqrt{48 x^{6}}-\sqrt[3]{125 x^{9}} \cdot 3^{\frac{1}{2}}}{\sqrt{75 x^{6}}}=\frac{124}{2} \times ? \): We start simplifying the numerator: \[ \sqrt{48x^6} = \sqrt{16 \cdot 3 \cdot x^6} = 4x^3\sqrt{3} \] Meanwhile, \[ \sqrt[3]{125x^9} = 5x^3 \quad\text{and thus, } 5x^3 \cdot 3^{\frac{1}{2}} = 5x^3\sqrt{3} \] Combining these in the numerator gives us: \[ 4x^3\sqrt{3} - 5x^3\sqrt{3} = -x^3\sqrt{3} \] Next, the denominator simplifies: \[ \sqrt{75x^6} = \sqrt{25 \cdot 3 \cdot x^6} = 5x^3\sqrt{3} \] Putting it all together, we get: \[ \frac{-x^3\sqrt{3}}{5x^3\sqrt{3}} = -\frac{1}{5} \] Now equating to \( \frac{124}{2} \times ? \) gives us the solution \( ? = -\frac{1}{5} \div 62 \), which leads to \( ? = -\frac{1}{310} \).