1. **State the problem:** Solve the inequality $$\frac{x-1}{x} - x \leq 1 - \frac{x}{x+1}$$. 2. **Rewrite the inequality:** Bring all terms to one side to compare with zero: $$\frac{x-1}{x} - x - 1 + \frac{x}{x+1} \leq 0$$. 3. **Find a common denominator:** The denominators are $x$ and $x+1$. The common denominator is $$x(x+1)$$. 4. **Rewrite each term with the common denominator:** - $$\frac{x-1}{x} = \frac{(x-1)(x+1)}{x(x+1)}$$ - $$-x = -\frac{x \cdot x(x+1)}{x(x+1)} = -\frac{x^2(x+1)}{x(x+1)}$$ - $$-1 = -\frac{x(x+1)}{x(x+1)}$$ - $$\frac{x}{x+1} = \frac{x \cdot x}{x(x+1)} = \frac{x^2}{x(x+1)}$$ 5. **Combine all terms over the common denominator:** $$\frac{(x-1)(x+1) - x^2(x+1) - x(x+1) + x^2}{x(x+1)} \leq 0$$ 6. **Simplify the numerator:** - Expand $(x-1)(x+1) = x^2 - 1$ - Expand $-x^2(x+1) = -x^3 - x^2$ - Expand $-x(x+1) = -x^2 - x$ - The last term is $+ x^2$ Sum numerator: $$x^2 - 1 - x^3 - x^2 - x^2 - x + x^2 = -x^3 - x - 1$$ 7. **Rewrite inequality:** $$\frac{-x^3 - x - 1}{x(x+1)} \leq 0$$ 8. **Analyze the inequality:** We want to find $x$ such that $$\frac{-x^3 - x - 1}{x(x+1)} \leq 0$$ 9. **Find critical points:** - Denominator zeros: $x=0$, $x=-1$ (excluded from domain) - Numerator zeros: Solve $$-x^3 - x - 1 = 0$$ or $$x^3 + x + 1 = 0$$ 10. **Solve cubic $x^3 + x + 1 = 0$:** By inspection, $x = -1$ gives $-1 + (-1) + 1 = -1 \neq 0$. Use approximate methods or note it has one real root approximately $x \approx -1.3247$. 11. **Sign analysis:** - Numerator changes sign at $x \approx -1.3247$ - Denominator changes sign at $x = -1$ and $x=0$ 12. **Test intervals:** - For $x < -1.3247$, numerator $>0$ (since cubic is increasing and root is at $-1.3247$) - For $-1.3247 < x < -1$, numerator $<0$ - For $-1 < x < 0$, denominator $<0$ (since $x$ and $x+1$ have opposite signs) - For $x > 0$, denominator $>0$ 13. **Determine where fraction is $\leq 0$:** Check signs on intervals: - $(-\infty, -1.3247)$: numerator $>0$, denominator $>0$ (since $x < -1$ and $x+1 < 0$, denominator $x(x+1) > 0$ because product of two negatives is positive), fraction $>0$ (not included) - $(-1.3247, -1)$: numerator $<0$, denominator $>0$, fraction $<0$ (included) - $(-1, 0)$: numerator $<0$, denominator $<0$, fraction $>0$ (not included) - $(0, \infty)$: numerator $<0$ (check at $x=1$: $-1 -1 -1 = -3 < 0$), denominator $>0$, fraction $<0$ (included) 14. **Domain restrictions:** Exclude $x = -1$ and $x=0$ because denominator zero. 15. **Final solution:** $$x \in (-1.3247, -1) \cup (0, \infty)$$ **Answer:** The solution to the inequality is $$\boxed{\left(-1.3247, -1\right) \cup (0, \infty)}$$.