1. **State the problem:** Simplify the expression $$\frac{2x^2 - 18}{x^2 - 4x + 4} \cdot \frac{x - 2}{2(x + 3)}$$ and simplify the expression $$\frac{x^2 - 1}{x^2 + 2x + 1} \div \frac{x - 1}{x + 1}$$ 2. **Recall important formulas and rules:** - Factor quadratic expressions when possible. - Division of fractions is multiplication by the reciprocal. - Cancel common factors in numerator and denominator. 3. **Simplify the first expression:** - Factor numerator and denominator: $$2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3)$$ $$x^2 - 4x + 4 = (x - 2)^2$$ - Rewrite the expression: $$\frac{2(x - 3)(x + 3)}{(x - 2)^2} \cdot \frac{x - 2}{2(x + 3)}$$ 4. **Multiply the fractions:** $$\frac{2(x - 3)(x + 3)}{(x - 2)^2} \times \frac{x - 2}{2(x + 3)} = \frac{2(x - 3)(x + 3)(x - 2)}{(x - 2)^2 \cdot 2(x + 3)}$$ 5. **Cancel common factors:** Cancel $2$: $$\frac{\cancel{2}(x - 3)(x + 3)(x - 2)}{(x - 2)^2 \cdot \cancel{2}(x + 3)} = \frac{(x - 3)(x + 3)(x - 2)}{(x - 2)^2 (x + 3)}$$ Cancel $(x + 3)$: $$\frac{(x - 3)\cancel{(x + 3)}(x - 2)}{(x - 2)^2 \cancel{(x + 3)}} = \frac{(x - 3)(x - 2)}{(x - 2)^2}$$ Cancel one $(x - 2)$: $$\frac{(x - 3)\cancel{(x - 2)}}{\cancel{(x - 2)}(x - 2)} = \frac{x - 3}{x - 2}$$ 6. **Simplify the second expression:** Rewrite division as multiplication by reciprocal: $$\frac{x^2 - 1}{x^2 + 2x + 1} \div \frac{x - 1}{x + 1} = \frac{x^2 - 1}{x^2 + 2x + 1} \times \frac{x + 1}{x - 1}$$ 7. **Factor expressions:** $$x^2 - 1 = (x - 1)(x + 1)$$ $$x^2 + 2x + 1 = (x + 1)^2$$ 8. **Substitute factors:** $$\frac{(x - 1)(x + 1)}{(x + 1)^2} \times \frac{x + 1}{x - 1}$$ 9. **Multiply numerators and denominators:** $$\frac{(x - 1)(x + 1)(x + 1)}{(x + 1)^2 (x - 1)}$$ 10. **Cancel common factors:** Cancel $(x - 1)$: $$\frac{\cancel{(x - 1)}(x + 1)(x + 1)}{(x + 1)^2 \cancel{(x - 1)}} = \frac{(x + 1)(x + 1)}{(x + 1)^2}$$ Cancel $(x + 1)^2$: $$\frac{\cancel{(x + 1)(x + 1)}}{\cancel{(x + 1)^2}} = 1$$ **Final answers:** $$\frac{x - 3}{x - 2}$$ and $$1$$