6. \( \frac{x+2}{x-1} \div \frac{x+4}{x^{2}+4 x-5} \) a. \( \frac{(x+2)(x+5)}{x+4}, x \neq-5,-4 \) c. \( \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x \neq 1,-5,-4 \) b. \( \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x \neq 1,-5 \) d. \( \frac{(x+2)(x+5)}{x+4}, x \neq 1,-4,-5 \) 7. \( \frac{x^{2}-4}{x-8} \div(x-2) \) a. \( \frac{(x-2)(x+2)}{x-8} \) c. \( \frac{x-2}{x-8} \) b. \( \frac{x-8}{x+2} \) d. \( \frac{x+2}{x-8} \) 8. \( \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 \) a. \( \quad-4 \) b. 3 or -4 c. \( \frac{-1 \pm \sqrt{73}}{2} \) d. 2 9. \( \frac{2}{3 m}+\frac{4}{5 m}=2 \) a. \( \frac{22}{15} \) b. \( \frac{13}{15} \) c. \( \frac{11}{15} \) d. \( \frac{3}{8} \) 10. \( \frac{5}{x+3}=\frac{-2}{x+4} \) a. \( -\frac{26}{5} \) b. \( -\frac{6}{7} \) c. \( -\frac{10}{7} \) d. \( -\frac{26}{7} \)

Let's solve each problem step by step. ### Problem 6: \[ \frac{x+2}{x-1} \div \frac{x+4}{x^{2}+4x-5} \] **Step 1: Rewrite the division as multiplication by the reciprocal.** \[ \frac{x+2}{x-1} \cdot \frac{x^{2}+4x-5}{x+4} \] **Step 2: Factor the quadratic expression \(x^{2}+4x-5\).** To factor \(x^{2}+4x-5\), we look for two numbers that multiply to \(-5\) and add to \(4\). These numbers are \(5\) and \(-1\). \[ x^{2}+4x-5 = (x+5)(x-1) \] **Step 3: Substitute the factored form back into the expression.** \[ \frac{x+2}{x-1} \cdot \frac{(x+5)(x-1)}{x+4} \] **Step 4: Cancel the common factor \((x-1)\).** \[ \frac{(x+2)(x+5)}{x+4} \] **Step 5: Identify restrictions.** The restrictions are \(x \neq -5\) and \(x \neq -4\) (from the denominator). **Final Answer for Problem 6:** \[ \frac{(x+2)(x+5)}{x+4}, \quad x \neq -5, -4 \] ### Problem 7: \[ \frac{x^{2}-4}{x-8} \div (x-2) \] **Step 1: Rewrite the division as multiplication by the reciprocal.** \[ \frac{x^{2}-4}{x-8} \cdot \frac{1}{x-2} \] **Step 2: Factor \(x^{2}-4\) (difference of squares).** \[ x^{2}-4 = (x-2)(x+2) \] **Step 3: Substitute the factored form back into the expression.** \[ \frac{(x-2)(x+2)}{x-8} \cdot \frac{1}{x-2} \] **Step 4: Cancel the common factor \((x-2)\).** \[ \frac{x+2}{x-8} \] **Final Answer for Problem 7:** \[ \frac{x+2}{x-8} \] ### Problem 8: \[ \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 \] **Step 1: Factor \(x^{2}-9\) (difference of squares).** \[ x^{2}-9 = (x-3)(x+3) \] **Step 2: Rewrite the equation.** \[ \frac{6}{(x-3)(x+3)} - \frac{1}{x-3} = 1 \] **Step 3: Find a common denominator, which is \((x-3)(x+3)\).** \[ \frac{6 - (x+3)}{(x-3)(x+3)} = 1 \] **Step 4: Simplify the numerator.** \[ \frac{6 - x - 3}{(x-3)(x+3)} = 1 \implies \frac{3 - x}{(x-3)(x+3)} = 1 \] **Step 5: Cross-multiply.** \[ 3 - x = (x-3)(x+3) \] **Step 6: Expand the right side.** \[ 3 - x = x^2 - 9 \] **Step 7: Rearrange the equation.** \[ x^2 + x - 12 = 0 \] **Step 8: Factor the quadratic.** \[ (x+4)(x-3) = 0 \] **Step 9: Solve for \(x\).** \[ x = -4 \quad \text{or} \quad x = 3 \] **Final Answer for Problem 8:** \[ 3 \text{ or } -4 \] ### Problem 9: \[ \frac{2}{3m}+\frac{4}{5m}=2 \] **Step 1: Find a common denominator, which is \(15m\).** \[ \frac{10}{15m} + \frac{12}{15m} = 2 \] **Step 2: Combine the fractions.** \[ \frac{22}{15m} = 2 \] **Step 3: Cross-multiply.** \[ 22 = 30m \] **Step 4: Solve for \(m\).** \[ m = \frac{22}{30} = \frac{11}{15} \] **Final Answer for Problem 9:** \[ \frac{11}{15} \] ### Problem 10: \[ \frac{5}{x+3}=\frac{-2}{x+4} \] **Step 1: Cross-multiply.** \[ 5(x+4) = -2(x+3) \] **Step 2: Expand both sides.** \[ 5x + 20 = -2x - 6 \] **Step 3: Rearrange the equation.** \[ 5x + 2x = -6 - 20 \] \[ 7x = -26 \] **Step 4: Solve for \(x\).** \[ x = -\frac{26}{7} \] **Final Answer for Problem 10:** \[ -\frac{26}{7} \] ### Summary of Answers: 6. \( \frac{(x+2)(x+5)}{x+4}, x \neq -5, -4 \) 7. \( \frac{x+2}{x-8} \) 8. \( 3 \text{ or } -4 \) 9. \( \frac{11}{15} \) 10. \( -\frac{26}{7} \)