1. **Stating the problem:** Prove that $$\sqrt{\frac{x}{1+c} + \sqrt{\frac{x}{1-c}}} = x^2 + \frac{3}{4} \cdot \frac{c^2}{x^2}$$ 2. **Understanding the expression:** We need to simplify the left-hand side (LHS) and show it equals the right-hand side (RHS). 3. **Rewrite the LHS:** Let $$A = \frac{x}{1+c} + \sqrt{\frac{x}{1-c}}$$ We want to find $$\sqrt{A}$$ 4. **Check the dimensions and terms:** The RHS has terms $x^2$ and $\frac{c^2}{x^2}$, so the LHS should simplify to a sum of these terms. 5. **Attempt to simplify the LHS:** Rewrite the nested square root: $$\sqrt{\frac{x}{1-c}} = \left(\frac{x}{1-c}\right)^{1/2}$$ 6. **Since the problem is likely a binomial approximation or expansion, consider small $c$ and expand denominators:** Use binomial expansion for denominators: $$\frac{1}{1+c} \approx 1 - c + c^2$$ $$\frac{1}{1-c} \approx 1 + c + c^2$$ 7. **Approximate each term:** $$\frac{x}{1+c} \approx x(1 - c + c^2) = x - x c + x c^2$$ $$\sqrt{\frac{x}{1-c}} = \sqrt{x(1 + c + c^2)} \approx \sqrt{x} \left(1 + \frac{c}{2} + \frac{3 c^2}{8}\right)$$ 8. **Sum inside the square root:** $$A \approx (x - x c + x c^2) + \sqrt{x} \left(1 + \frac{c}{2} + \frac{3 c^2}{8}\right)$$ 9. **Since the RHS is $x^2 + \frac{3}{4} \frac{c^2}{x^2}$, the problem likely has a typo or requires a different approach.** 10. **Alternative approach: Square both sides and verify equality:** Let $$y = \sqrt{\frac{x}{1+c} + \sqrt{\frac{x}{1-c}}}$$ Then $$y^2 = \frac{x}{1+c} + \sqrt{\frac{x}{1-c}}$$ 11. **Try to express RHS in terms of $x$ and $c$ and check if $y^2$ matches $\left(x^2 + \frac{3}{4} \frac{c^2}{x^2}\right)^2$.** 12. **Since the problem as stated is ambiguous or incomplete, the best we can do is note the difficulty and suggest verifying the problem statement.** **Final note:** The given expression does not simplify directly to the RHS as stated without additional context or corrections. **Hence, the proof cannot be completed as given.**