1. Problem: Simplify $$\frac{2}{x+1} + \frac{1}{x-2} - \frac{1}{x^2 - 1}$$. 2. Note that $$x^2 - 1 = (x+1)(x-1)$$. 3. Find common denominator: $$ (x+1)(x-2)(x-1) $$. 4. Rewrite each fraction with the common denominator: $$\frac{2}{x+1} = \frac{2(x-2)(x-1)}{(x+1)(x-2)(x-1)}$$ $$\frac{1}{x-2} = \frac{1(x+1)(x-1)}{(x+1)(x-2)(x-1)}$$ $$\frac{1}{x^2 -1} = \frac{1(x-2)}{(x+1)(x-2)(x-1)}$$ 5. Combine numerators: $$2(x-2)(x-1) + (x+1)(x-1) - (x-2)$$ 6. Expand terms: $$2(x^2 -3x + 2) + (x^2 -1) - (x - 2)$$ $$= 2x^2 -6x +4 + x^2 -1 - x + 2$$ 7. Combine like terms: $$3x^2 -7x + 5$$ 8. So the simplified expression is: $$\frac{3x^2 - 7x + 5}{(x+1)(x-2)(x-1)} = \frac{3x^2 - 7x + 5}{(x - 2)(x^2 - 1)}$$ Answer: (c) --- 1. Problem: Simplify $$\frac{1}{x-3} + \frac{1}{x+3}$$. 2. Common denominator: $$ (x-3)(x+3) = x^2 - 9 $$. 3. Rewrite fractions: $$\frac{1(x+3)}{x^2-9} + \frac{1(x-3)}{x^2-9} = \frac{x+3+x-3}{x^2-9} = \frac{2x}{x^2-9}$$ Answer: (b) --- 1. Problem: Simplify $$-\frac{2}{x} + \frac{1}{x-1} + \frac{1}{x+1}$$. 2. Find common denominator: $$x(x-1)(x+1) = x(x^2 - 1)$$. 3. Rewrite fractions with common denominator: $$-\frac{2(x-1)(x+1)}{x(x-1)(x+1)} + \frac{1(x)(x+1)}{x(x-1)(x+1)} + \frac{1(x)(x-1)}{x(x-1)(x+1)}$$ 4. Expand numerators: $$-2(x^2 - 1) + x(x+1) + x(x-1) = -2x^2 + 2 + x^2 + x + x^2 - x = 2$$ 5. Simplify numerator: $$ -2x^2 + 2 + x^2 + x + x^2 - x = 2 $$ 6. So expression is: $$\frac{2}{x(x^2 - 1)}$$ Answer: (c) --- 1. Problem: Simplify $$\frac{x^2 - 3x + 2}{x^2 + 5x + 4} \times \frac{x + 1}{x - 1}$$. 2. Factor polynomials: $$x^2 - 3x + 2 = (x - 1)(x - 2)$$ $$x^2 + 5x + 4 = (x + 4)(x + 1)$$ 3. Substitute factors: $$\frac{(x - 1)(x - 2)}{(x + 4)(x + 1)} \times \frac{x + 1}{x - 1}$$ 4. Cancel common terms: $$\frac{(x - 2)}{(x + 4)}$$ Answer: (c) --- 1. Problem: Simplify $$\frac{4x^2 - 4}{x^2 + 2x + 1} \cdot \frac{x + 1}{x - 1}$$. 2. Factor: $$4x^2 - 4 = 4(x^2 - 1) = 4(x-1)(x+1)$$ $$x^2 + 2x + 1 = (x + 1)^2$$ 3. Substitute: $$\frac{4(x-1)(x+1)}{(x + 1)^2} \times \frac{x + 1}{x - 1}$$ 4. Cancel terms: Cancel $$ (x+1) $$ and $$ (x-1) $$ in numerator and denominator: $$\frac{4(x+1)(x-1)}{(x + 1)^2} \times \frac{x + 1}{x - 1} = 4$$ Answer: (c) --- 1. Problem: Simplify $$\frac{x^2 - 9}{x^2 + 6x + 9} \cdot \frac{x + 3}{x - 9}$$. 2. Factor: $$x^2 - 9 = (x - 3)(x + 3)$$ $$x^2 + 6x + 9 = (x + 3)^2$$ 3. Substitute: $$\frac{(x - 3)(x + 3)}{(x + 3)^2} \times \frac{x + 3}{x - 9}$$ 4. Cancel $$ (x + 3) $$: $$\frac{x - 3}{x + 3} \times \frac{1}{x - 9} (x + 3)$$ , reduces to $$\frac{x - 3}{x - 9}$$ Answer: (a) --- 1. Problem: Simplify $$\frac{3x^2 - 12}{x + 2} \cdot \frac{x}{x - 2}$$. 2. Factor numerator: $$3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$$ 3. Substitute: $$\frac{3(x-2)(x+2)}{x + 2} \cdot \frac{x}{x - 2}$$ 4. Cancel $$ (x + 2) $$ and $$ (x - 2) $$: $$3x$$ Answer: (a) --- 1. Problem: Simplify $$\frac{2 - 3x - 2x^2}{x^2 + 3x} \div \frac{x^2 + 3x + 2}{x + 3}$$. 2. Factor denominators and numerators: Numerator of first fraction: $$2 - 3x - 2x^2 = - (2x^2 + 3x - 2)$$ Factor: $$2x^2 + 3x - 2 = (2x - 1)(x + 2)$$ So numerator is $$- (2x - 1)(x + 2)$$ 3. Denominator of first fraction: $$x^2 + 3x = x(x + 3)$$ 4. Numerator of second fraction: $$x^2 + 3x + 2 = (x + 1)(x + 2)$$ 5. Rewrite division as multiplication by reciprocal: $$\frac{- (2x - 1)(x + 2)}{x(x + 3)} \times \frac{x + 3}{(x + 1)(x + 2)}$$ 6. Cancel $$ (x + 2) $$ and $$ (x + 3) $$: $$\frac{-(2x - 1)}{x} \times \frac{1}{x + 1} = \frac{-(2x - 1)}{x(x + 1)} = \frac{1 - 2x}{x(x + 1)}$$ Answer: (c) --- 1. Problem: Simplify $$\frac{x + 2}{1 + \frac{2}{x}}$$. 2. Simplify denominator: $$1 + \frac{2}{x} = \frac{x + 2}{x}$$ 3. Therefore: $$\frac{x + 2}{\frac{x + 2}{x}} = (x + 2) \times \frac{x}{x + 2} = x$$ Answer: (a)