d) \( \frac{x^{2}+x-6}{3 x^{2}-12 x} \div \frac{x^{3}-2 x^{2}}{x^{2}-16} \times \frac{1}{x+4} \)

To simplify the given expression \( \frac{x^{2}+x-6}{3 x^{2}-12 x} \div \frac{x^{3}-2 x^{2}}{x^{2}-16} \times \frac{1}{x+4} \), we can follow these steps: First, we'll factor each part of the expression: 1. **Numerator of the first fraction:** \( x^2 + x - 6 \) can be factored as \( (x + 3)(x - 2) \). 2. **Denominator of the first fraction:** \( 3x^2 - 12x = 3x(x - 4) \). 3. **Numerator of the second fraction:** \( x^3 - 2x^2 = x^2(x - 2) \). 4. **Denominator of the second fraction:** \( x^2 - 16 = (x - 4)(x + 4) \) (this is a difference of squares). Now, substitute back the factored forms into the original problem: \[ \frac{(x + 3)(x - 2)}{3x(x - 4)} \div \frac{x^2(x - 2)}{(x - 4)(x + 4)} \times \frac{1}{x + 4} \] Changing the division to multiplication by the reciprocal: \[ \frac{(x + 3)(x - 2)}{3x(x - 4)} \times \frac{(x - 4)(x + 4)}{x^2(x - 2)} \times \frac{1}{x + 4} \] Next, we'll cancel out common factors: - The \( (x - 2) \) in the numerator and denominator. - The \( (x - 4) \) terms cancel out. Our current expression simplifies to: \[ \frac{(x + 3)}{3x} \times \frac{(x + 4)}{x} \times \frac{1}{x + 4} \] The \( (x + 4) \) also cancels, leading to: \[ \frac{(x + 3)}{3x \cdot x} = \frac{x + 3}{3x^2} \] Thus, the final simplified expression is: \[ \frac{x + 3}{3x^2} \]