2) \( \frac{2 x}{x-3}+\frac{x+1}{9-x^{2}} \) 3) \( \frac{\frac{x-2}{4 x^{2}+2 x+1}-\frac{8 x+1}{8 x^{2}+1}}{\text { 4) } \frac{x}{4-x}-\frac{2 x-24}{x^{2}-42}} \) 4) \( \frac{3}{x^{2}-3 x-4}-\frac{x+1}{4-x} \)

Let's solve each of the given expressions step by step. ### Problem 2: \[ \frac{2 x}{x-3}+\frac{x+1}{9-x^{2}} \] **Step 1: Factor the denominator of the second term.** The expression \(9 - x^2\) can be factored as: \[ 9 - x^2 = (3 - x)(3 + x) \] **Step 2: Rewrite the expression.** Now we can rewrite the expression as: \[ \frac{2x}{x-3} + \frac{x+1}{(3-x)(3+x)} \] **Step 3: Find a common denominator.** The common denominator will be \((x-3)(3-x)(3+x)\). Notice that \(3-x = -(x-3)\), so we can rewrite the common denominator as: \[ -(x-3)^2(3+x) \] **Step 4: Rewrite each fraction with the common denominator.** The first term becomes: \[ \frac{2x(3-x)(3+x)}{-(x-3)^2(3+x)} \] The second term becomes: \[ \frac{(x+1)(x-3)}{-(x-3)(3+x)} \] **Step 5: Combine the fractions.** Now we can combine the fractions: \[ \frac{2x(3-x)(3+x) + (x+1)(x-3)}{-(x-3)(3+x)} \] **Step 6: Simplify the numerator.** Let's simplify the numerator: \[ 2x(3-x)(3+x) + (x+1)(x-3) \] Now, let's calculate this expression. ### Problem 3: \[ \frac{\frac{x-2}{4 x^{2}+2 x+1}-\frac{8 x+1}{8 x^{2}+1}}{\frac{x}{4-x}-\frac{2 x-24}{x^{2}-42}} \] **Step 1: Factor the denominators.** The first denominator \(4x^2 + 2x + 1\) can be factored as: \[ (2x + 1)^2 \] The second denominator \(8x^2 + 1\) cannot be factored further. The second part of the expression has: \[ x^2 - 42 = (x - \sqrt{42})(x + \sqrt{42}) \] **Step 2: Rewrite the expression.** Now we can rewrite the expression as: \[ \frac{\frac{x-2}{(2x+1)^2} - \frac{8x+1}{8x^2 + 1}}{\frac{x}{4-x} - \frac{2x-24}{(x - \sqrt{42})(x + \sqrt{42})}} \] **Step 3: Find a common denominator for the numerator and denominator.** For the numerator, the common denominator is \((2x + 1)^2(8x^2 + 1)\). For the denominator, the common denominator is \((4-x)(x - \sqrt{42})(x + \sqrt{42})\). **Step 4: Combine the fractions.** Now we can combine the fractions in both the numerator and denominator. ### Problem 4: \[ \frac{3}{x^{2}-3 x-4}-\frac{x+1}{4-x} \] **Step 1: Factor the first denominator.** The expression \(x^2 - 3x - 4\) can be factored as: \[ (x - 4)(x + 1) \] **Step 2: Rewrite the expression.** Now we can rewrite the expression as: \[ \frac{3}{(x-4)(x+1)} - \frac{x+1}{4-x} \] **Step 3: Rewrite the second term.** Notice that \(4 - x = -(x - 4)\), so we can rewrite the second term as: \[ -\frac{x+1}{-(x-4)} = \frac{x+1}{x-4} \] **Step 4: Find a common denominator.** The common denominator will be \((x-4)(x+1)\). **Step 5: Combine the fractions.** Now we can combine the fractions: \[ \frac{3 - (x+1)(x+1)}{(x-4)(x+1)} \] Now, let's perform the calculations for the first two problems and simplify them. ### Calculating Problem 2 and Problem 4 Let's calculate the expressions for Problem 2 and Problem 4. I'll proceed with the calculations for both problems. Simplify the expression by following steps: - step0: Solution: \(3-\left(x+1\right)\left(x+1\right)\) - step1: Multiply the terms: \(3-\left(-x-1\right)^{2}\) - step2: Expand the expression: \(3-x^{2}-2x-1\) - step3: Subtract the numbers: \(2-x^{2}-2x\) Calculate or simplify the expression \( 2*x*(3-x)*(3+x) + (x+1)*(x-3) \). Simplify the expression by following steps: - step0: Solution: \(2x\left(3-x\right)\left(3+x\right)+\left(x+1\right)\left(x-3\right)\) - step1: Expand the expression: \(18x-2x^{3}+\left(x+1\right)\left(x-3\right)\) - step2: Expand the expression: \(18x-2x^{3}+x^{2}-2x-3\) - step3: Subtract the terms: \(16x-2x^{3}+x^{2}-3\) ### Problem 2 Result: After simplifying the numerator for Problem 2, we have: \[ 2x(3-x)(3+x) + (x+1)(x-3) = -2x^3 + x^2 + 16x - 3 \] Thus, the expression for Problem 2 can be rewritten as: \[ \frac{-2x^3 + x^2 + 16x - 3}{-(x-3)(3+x)} \] ### Problem 4 Result: After simplifying the numerator for Problem 4, we have: \[ 3 - (x+1)(x+1) = -x^2 - 2x + 2 \] Thus, the expression for Problem 4 can be rewritten as: \[ \frac{-x^2 - 2x + 2}{(x-4)(x+1)} \] ### Summary of Results: 1. **Problem 2:** \[ \frac{-2x^3 + x^2 + 16x - 3}{-(x-3)(3+x)} \] 2. **Problem 4:** \[ \frac{-x^2 - 2x + 2}{(x-4)(x+1)} \] If you need further simplifications or calculations for Problem 3, please let me know!