1. **State the problem:** Solve the equation $$\frac{1}{1 - x} + \frac{1}{1 + \sqrt{x}} = \frac{1}{1 - \sqrt{x}}$$ for $x$. 2. **Identify the domain:** Since $\sqrt{x}$ appears, $x \geq 0$. Also, denominators cannot be zero, so: - $1 - x \neq 0 \Rightarrow x \neq 1$ - $1 + \sqrt{x} \neq 0 \Rightarrow \sqrt{x} \neq -1$ (always true since $\sqrt{x} \geq 0$) - $1 - \sqrt{x} \neq 0 \Rightarrow \sqrt{x} \neq 1 \Rightarrow x \neq 1$ 3. **Rewrite the equation:** Let $y = \sqrt{x}$, so $x = y^2$ and $y \geq 0$. The equation becomes: $$\frac{1}{1 - y^2} + \frac{1}{1 + y} = \frac{1}{1 - y}$$ 4. **Find a common denominator:** The denominators are $1 - y^2 = (1 - y)(1 + y)$, $1 + y$, and $1 - y$. The least common denominator (LCD) is $(1 - y)(1 + y)$. 5. **Rewrite each term with the LCD:** $$\frac{1}{(1 - y)(1 + y)} + \frac{1}{1 + y} = \frac{1}{1 - y}$$ Multiply numerator and denominator of the second term by $(1 - y)$: $$\frac{1}{(1 - y)(1 + y)} + \frac{1 \cdot (1 - y)}{(1 + y)(1 - y)} = \frac{1}{1 - y}$$ 6. **Combine the left side:** $$\frac{1 + (1 - y)}{(1 - y)(1 + y)} = \frac{1}{1 - y}$$ Simplify numerator: $$1 + (1 - y) = 2 - y$$ So: $$\frac{2 - y}{(1 - y)(1 + y)} = \frac{1}{1 - y}$$ 7. **Cross-multiply:** $$(2 - y)(1 - y) = (1)(1 + y)$$ 8. **Expand both sides:** $$(2 - y)(1 - y) = 2(1 - y) - y(1 - y) = 2 - 2y - y + y^2 = 2 - 3y + y^2$$ Right side: $$1 + y$$ So the equation is: $$2 - 3y + y^2 = 1 + y$$ 9. **Bring all terms to one side:** $$y^2 - 3y + 2 - 1 - y = 0 \Rightarrow y^2 - 4y + 1 = 0$$ 10. **Solve quadratic equation:** $$y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$ 11. **Check domain $y \geq 0$:** Both $2 + \sqrt{3} > 0$ and $2 - \sqrt{3} > 0$ (since $\sqrt{3} \approx 1.732$). 12. **Check denominators for $y = 2 + \sqrt{3}$:** - $1 - y = 1 - (2 + \sqrt{3}) < 0$ but not zero, valid. - $1 + y > 0$, valid. - $1 - y^2 = 1 - (2 + \sqrt{3})^2 \neq 0$, valid. 13. **Check denominators for $y = 2 - \sqrt{3}$:** - $1 - y = 1 - (2 - \sqrt{3}) = (1 - 2) + \sqrt{3} = -1 + 1.732 = 0.732 \neq 0$, valid. - $1 + y > 0$, valid. - $1 - y^2 = 1 - (2 - \sqrt{3})^2 \neq 0$, valid. 14. **Convert back to $x$:** $$x = y^2$$ For $y = 2 + \sqrt{3}$: $$x = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$$ For $y = 2 - \sqrt{3}$: $$x = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$$ 15. **Final answer:** $$x = 7 + 4\sqrt{3} \quad \text{or} \quad x = 7 - 4\sqrt{3}$$