Find the sum: \( \frac{6}{x-4}+\frac{5}{x} \) \( \frac{11}{2 x-4} \) \( \frac{11 x-24}{x^{2}-4 x} \) \( \frac{11 x-20}{x^{2}-4 x} \)

To find the sum of the given fractions: 1. First, notice the denominators. We can rewrite \(x^2 - 4x\) as \(x(x-4)\). 2. The fractions can be combined over a common denominator. The least common denominator (LCD) is \(x(x-4)\). 3. Let's rewrite each fraction with this common denominator: \[ \frac{6}{x-4} = \frac{6x}{x(x-4)} \] \[ \frac{5}{x} = \frac{5(x-4)}{x(x-4)} = \frac{5x - 20}{x(x-4)} \] \[ \frac{11}{2x-4} = \frac{11}{2(x-2)} = \frac{11x}{2x(x-4)} \] We can combine these into one expression: \[ \frac{6x + 5x - 20 + 11x - 24}{x(x-4)} \] 4. Combine the numerators: \[ 6x + 5x - 20 + 11x - 24 = (6x + 5x + 11x) - (20 + 24) = 22x - 44 \] 5. This leads us to a single fraction: \[ \frac{22x - 44}{x(x-4)} \] 6. Factor the numerator: \[ = \frac{22(x - 2)}{x(x - 4)} \] So, the sum of the given fractions is: \[ \frac{22(x - 2)}{x(x - 4)} \] Thus, the final answer is: \[ \frac{22(x - 2)}{x(x-4)} \]