1. **Simplify the expression** $\frac{n}{5-n} + \frac{2n-5}{n-5}$. Note that $n-5 = -(5-n)$, so rewrite the second fraction: $$\frac{2n-5}{n-5} = \frac{2n-5}{-(5-n)} = -\frac{2n-5}{5-n}$$ Now the expression becomes: $$\frac{n}{5-n} - \frac{2n-5}{5-n} = \frac{n - (2n-5)}{5-n} = \frac{n - 2n + 5}{5-n} = \frac{5 - n}{5-n}$$ Since $5-n$ is the numerator and denominator, the expression simplifies to: $$-1$$ --- 2. **Solve the equation** $|5x - 3| = -52$. The absolute value $|5x - 3|$ is always non-negative, so it cannot equal a negative number like $-52$. Therefore, the solution set is: $$\emptyset$$ (the empty set). --- 3. **Find values of $k$ such that** $f(x) = \frac{x}{x^2 + k}$ has an extreme value at $x=2$. Step 1: Find the derivative $f'(x)$ using the quotient rule: $$f'(x) = \frac{(1)(x^2 + k) - x(2x)}{(x^2 + k)^2} = \frac{x^2 + k - 2x^2}{(x^2 + k)^2} = \frac{k - x^2}{(x^2 + k)^2}$$ Step 2: Set $f'(2) = 0$ for an extreme value at $x=2$: $$\frac{k - (2)^2}{(2^2 + k)^2} = 0 \implies k - 4 = 0 \implies k = 4$$ **Final answers:** 1. Simplified expression: $-1$ 2. Solution set of $|5x - 3| = -52$: $\emptyset$ 3. Value of $k$ for extreme value at $x=2$: $4$