Activity 10 Simplify the following \( \frac{\text { expressions }}{\text { a) } \frac{x^{2}+5 x+6}{x^{2}-9} \div \frac{x^{2}-4}{x^{2}-5 x+6}} \) b) \( \frac{x}{x-2}-\frac{1}{x+2}+\frac{8}{4-x^{2}} \)

To simplify the expression given in part a), we first handle the division of the two fractions. Start with: \[ \frac{x^{2}+5 x+6}{x^{2}-9} \div \frac{x^{2}-4}{x^{2}-5 x+6} \] This division can be converted to multiplication by taking the reciprocal of the second fraction: \[ \frac{x^{2}+5 x+6}{x^{2}-9} \times \frac{x^{2}-5 x+6}{x^{2}-4} \] Next, we factor the polynomials: - \(x^{2}+5x+6 = (x+2)(x+3)\) - \(x^{2}-9 = (x-3)(x+3)\) - \(x^{2}-4 = (x-2)(x+2)\) - \(x^{2}-5x+6 = (x-2)(x-3)\) Now substitute these factors into the expression: \[ \frac{(x+2)(x+3)}{(x-3)(x+3)} \times \frac{(x-2)(x-3)}{(x-2)(x+2)} \] Now, we can cancel common factors: - \(x+2\) cancels with \(x+2\), - \(x-2\) cancels with \(x-2\), - \(x+3\) cancels with \(x+3\), - and \(x-3\) cancels with \(x-3\). After simplification, we're left with: \[ 1 \] ### For part b): The expression is: \[ \frac{x}{x-2} - \frac{1}{x+2} + \frac{8}{4-x^{2}} \] First, note that \(4-x^{2}\) can be factored as \((2-x)(2+x)\): \[ \frac{x}{x-2} - \frac{1}{x+2} + \frac{8}{(2-x)(2+x)} \] To combine the fractions, we find a common denominator which will be \((x-2)(x+2)(2-x)(2+x)\). We rewrite each fraction with this common denominator and then combine: 1. For the first term: \[ \frac{x(2+x)}{(x-2)(x+2)(2-x)(2+x)} \] 2. For the second term: \[ -\frac{(2-x)(2+x)}{(x-2)(x+2)(2-x)(2+x)} \] 3. For the third term: \[ \frac{8(x-2)}{(x-2)(x+2)(2-x)(2+x)} \] Bringing these together gives: \[ \frac{x(2+x) - (2-x)(2+x) + 8(x-2)}{(x-2)(x+2)(2-x)(2+x)} \] After simplifying the numerator, collect like terms, leading to: \[ \frac{(x^2 + 2x) - (4 - x^2) + (8x - 16)}{(x-2)(x+2)(2-x)(2+x)} \] Combine and rearrange the numerator, then simplify to find the final answer.