Simplify. \[ \frac{\frac{3 x+21}{28 x^{4}}}{\frac{5 x+35}{4 x^{2}}} \]

To simplify the expression \[ \frac{\frac{3 x+21}{28 x^{4}}}{\frac{5 x+35}{4 x^{2}}} \] we will first simplify both the numerator and the denominator separately. **Step 1: Simplifying the Numerator** The numerator is \[ 3x + 21 \] Factoring this gives: \[ 3(x + 7). \] So the numerator becomes \[ \frac{3(x + 7)}{28x^4}. \] **Step 2: Simplifying the Denominator** The denominator is \[ 5x + 35. \] Factoring this gives: \[ 5(x + 7). \] Thus, the denominator becomes \[ \frac{5(x + 7)}{4x^2}. \] **Step 3: Substituting back into the expression** The expression now looks like: \[ \frac{\frac{3(x + 7)}{28x^4}}{\frac{5(x + 7)}{4x^2}}. \] **Step 4: Dividing fractions** To divide these fractions, we multiply by the reciprocal of the denominator: \[ \frac{3(x + 7)}{28x^4} \cdot \frac{4x^2}{5(x + 7)}. \] Now we can cancel \( (x + 7) \) from the numerator and the denominator: \[ = \frac{3 \cdot 4x^2}{28x^4 \cdot 5}. \] **Step 5: Simplifying further** This simplifies to: \[ = \frac{12x^2}{140x^4}. \] Now we can simplify the coefficients and the powers of \( x \): \[ \frac{12}{140} = \frac{6}{70} = \frac{3}{35}, \] and \[ \frac{x^2}{x^4} = \frac{1}{x^{2}}. \] Thus, our expression reduces to: \[ \frac{3}{35x^2}. \] **Final Answer:** The simplified expression is \[ \boxed{\frac{3}{35x^2}}. \]