1. **State the problem:** Solve the equation $$\frac{x^2 + 25 + 10x}{x^2 + 2x - 15} \left( \frac{x}{x+5} - \frac{x}{x+2} : \frac{x+1}{x+2} \right) = 0$$ 2. **Rewrite and simplify the expression inside the parentheses:** Recall that division of fractions means multiplying by the reciprocal: $$\frac{x}{x+5} - \frac{x}{x+2} : \frac{x+1}{x+2} = \frac{x}{x+5} - \frac{x}{x+2} \times \frac{x+2}{x+1}$$ 3. **Simplify the division part:** $$\frac{x}{x+2} : \frac{x+1}{x+2} = \frac{x}{x+2} \times \frac{x+2}{x+1} = \frac{x}{x+1}$$ 4. **Rewrite the parentheses:** $$\frac{x}{x+5} - \frac{x}{x+1}$$ 5. **Find common denominator and subtract:** $$\frac{x(x+1)}{(x+5)(x+1)} - \frac{x(x+5)}{(x+1)(x+5)} = \frac{x(x+1) - x(x+5)}{(x+5)(x+1)}$$ 6. **Simplify numerator:** $$x(x+1) - x(x+5) = x^2 + x - x^2 - 5x = -4x$$ 7. **So the parentheses simplify to:** $$\frac{-4x}{(x+5)(x+1)}$$ 8. **Rewrite the original equation:** $$\frac{x^2 + 25 + 10x}{x^2 + 2x - 15} \times \frac{-4x}{(x+5)(x+1)} = 0$$ 9. **Factor where possible:** - Numerator of first fraction: $$x^2 + 10x + 25 = (x+5)^2$$ - Denominator of first fraction: $$x^2 + 2x - 15 = (x+5)(x-3)$$ 10. **Substitute factorizations:** $$\frac{(x+5)^2}{(x+5)(x-3)} \times \frac{-4x}{(x+5)(x+1)} = 0$$ 11. **Cancel common factors:** $$\frac{\cancel{(x+5)}(x+5)}{\cancel{(x+5)}(x-3)} \times \frac{-4x}{(x+5)(x+1)} = \frac{(x+5)}{(x-3)} \times \frac{-4x}{(x+5)(x+1)}$$ 12. **Cancel another (x+5):** $$\frac{\cancel{(x+5)}}{(x-3)} \times \frac{-4x}{\cancel{(x+5)}(x+1)} = \frac{-4x}{(x-3)(x+1)}$$ 13. **Final simplified equation:** $$\frac{-4x}{(x-3)(x+1)} = 0$$ 14. **Solve for x:** A fraction equals zero only if its numerator is zero (and denominator is not zero): $$-4x = 0 \implies x = 0$$ 15. **Check for restrictions:** Denominator cannot be zero: $$x-3 \neq 0 \implies x \neq 3$$ $$x+1 \neq 0 \implies x \neq -1$$ $$x+5 \neq 0 \implies x \neq -5$$ Since $x=0$ does not violate any restrictions, it is the solution. **Final answer:** $$x = 0$$