1. Stating the problems: - Problem 1: Solve the equation $$\frac{1}{x} + \frac{1}{x+1} = \frac{1}{x(x+1)}$$. - Problem 2: Solve the equation $$\frac{1}{x+2} - \frac{4}{x-2} = \frac{2x}{x^2 - 4}$$. - Problem 3: Solve the equation $$2\left(\frac{x}{x-3}\right)^2 + \frac{5x}{3-x} + 3 = 0$$. 2. Problem 1 solution: - The denominators are $x$, $x+1$, and $x(x+1)$. - Multiply both sides by $x(x+1)$ to clear denominators: $$x(x+1)\left(\frac{1}{x} + \frac{1}{x+1}\right) = x(x+1) \cdot \frac{1}{x(x+1)}$$ - Simplify left side: $$ (x+1) + x = 1$$ - Combine like terms: $$2x + 1 = 1$$ - Subtract 1 from both sides: $$2x = 0$$ - Divide both sides by 2: $$x = 0$$ - Check for restrictions: denominators cannot be zero, so $x \neq 0$ and $x \neq -1$. - Since $x=0$ is restricted, no solution for problem 1. 3. Problem 2 solution: - Note that $x^2 - 4 = (x-2)(x+2)$. - Multiply both sides by $x^2 - 4$: $$ (x^2 - 4) \left( \frac{1}{x+2} - \frac{4}{x-2} \right) = (x^2 - 4) \cdot \frac{2x}{x^2 - 4}$$ - Simplify left side: $$ (x-2) - 4(x+2) = 2x$$ - Expand right side: $$x - 2 - 4x - 8 = 2x$$ - Combine like terms on left: $$-3x - 10 = 2x$$ - Add $3x$ to both sides: $$-10 = 5x$$ - Divide by 5: $$x = -2$$ - Restrictions: denominators require $x \neq -2$ and $x \neq 2$. - $x = -2$ is excluded, so no solution for problem 2. 4. Problem 3 solution: - Note $3 - x = -(x - 3)$ so $$\frac{5x}{3-x} = \frac{5x}{-(x-3)} = -\frac{5x}{x-3}$$. - Rewrite equation: $$2\left(\frac{x}{x-3}\right)^2 - \frac{5x}{x-3} + 3 = 0$$ - Let $$y = \frac{x}{x-3}$$, so equation becomes: $$2y^2 - 5y + 3 = 0$$ - Solve quadratic in $y$ using quadratic formula: $$y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}$$ - Two solutions for $y$: $$y_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$$ $$y_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$$ - Recall $$y = \frac{x}{x-3}$$, so solve: For $y_1 = \frac{3}{2}$: $$\frac{x}{x-3} = \frac{3}{2} \Rightarrow 2x = 3(x - 3) \Rightarrow 2x = 3x - 9 \Rightarrow -x = -9 \Rightarrow x = 9$$ For $y_2 = 1$: $$\frac{x}{x-3} = 1 \Rightarrow x = x - 3$$ No solution from second equation. - Restrictions: denominator $x - 3 \neq 0 \Rightarrow x \neq 3$. - Final solution is $x=9$. Final answers: - Problem 1: No valid solution. - Problem 2: No valid solution. - Problem 3: $x=9$.