(3) Simplify: \( \frac{\frac{1}{4}-\frac{1}{8 x}}{\frac{1}{x}-\frac{1}{2 x^{2}}} \)

To simplify the expression \[ \frac{\frac{1}{4}-\frac{1}{8 x}}{\frac{1}{x}-\frac{1}{2 x^{2}}} \] we will start by simplifying both the numerator and the denominator separately. **Step 1: Simplify the numerator** The numerator is \[ \frac{1}{4} - \frac{1}{8x}. \] To combine these fractions, we need a common denominator. The least common multiple of \(4\) and \(8x\) is \(8x\). We rewrite each term with this common denominator: \[ \frac{1}{4} = \frac{2x}{8x} \] Thus, we have: \[ \frac{1}{4} - \frac{1}{8x} = \frac{2x}{8x} - \frac{1}{8x} = \frac{2x - 1}{8x}. \] **Step 2: Simplify the denominator** The denominator is \[ \frac{1}{x} - \frac{1}{2x^2}. \] Again, we need a common denominator. The least common multiple of \(x\) and \(2x^2\) is \(2x^2\). We rewrite each term: \[ \frac{1}{x} = \frac{2}{2x^2} \] Thus, we have: \[ \frac{1}{x} - \frac{1}{2x^2} = \frac{2}{2x^2} - \frac{1}{2x^2} = \frac{2 - 1}{2x^2} = \frac{1}{2x^2}. \] **Step 3: Combine the simplified numerator and denominator** Now we can substitute the simplified forms back into the original expression: \[ \frac{\frac{2x - 1}{8x}}{\frac{1}{2x^2}}. \] Dividing by a fraction is the same as multiplying by its reciprocal: \[ \frac{2x - 1}{8x} \cdot \frac{2x^2}{1} = \frac{(2x - 1) \cdot 2x^2}{8x}. \] **Step 4: Simplify the expression** Now we simplify: \[ \frac{(2x - 1) \cdot 2x^2}{8x} = \frac{2(2x - 1)x^2}{8x}. \] We can cancel \(x\) in the numerator and denominator (assuming \(x \neq 0\)): \[ = \frac{2(2x - 1)x}{8} = \frac{(2x - 1)x}{4}. \] Thus, the simplified expression is \[ \frac{(2x - 1)x}{4}. \] Therefore, the final answer is \[ \boxed{\frac{(2x - 1)x}{4}}. \]