To solve the equation \(\sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = \frac{25}{12}\), we need to find the value of \(x\) that satisfies this equation. Let's solve it step-by-step.

1. Simplify the expression using the substitution: \(a = \sqrt{\frac{x}{1-x}}\) and \(b = \sqrt{\frac{1-x}{x}}\).
2. From these, we have two equations: \(a + b = \frac{25}{12}\) and \(ab = 1\) (since \(a^2 \cdot b^2 = \frac{x}{1-x} \cdot \frac{1-x}{x} = 1\)).
3. Now use the identity: \((a + b)^2 = a^2 + b^2 + 2ab\). Substituting the known values, we get:
   • \((a + b)^2 = \left(\frac{25}{12}\right)^2 = \frac{625}{144}\)
   • \(a^2 + b^2 = \frac{625}{144} - 2 \times 1 = \frac{625}{144} - \frac{288}{144} = \frac{337}{144}\)
4. Since \(a^2 = \frac{x}{1-x}\) and \(b^2 = \frac{1-x}{x}\), from \(ab = 1\), we have:
   • \(\frac{x}{1-x} + \frac{1-x}{x} = \frac{337}{144}\)
   • Multiply both sides by \(x(1-x)\) to eliminate denominators: \(x^2 + (1-x)^2 = \frac{337}{144} \cdot x(1-x)\)
5. Solving this quadratic equation gives possible values for \(x\). Let's test the options:
   • For \(x = \frac{9}{25}\):
     • \(\sqrt{\frac{x}{1-x}} = \sqrt{\frac{\frac{9}{25}}{\frac{16}{25}}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\)
     • \(\sqrt{\frac{1-x}{x}} = \sqrt{\frac{\frac{16}{25}}{\frac{9}{25}}} = \sqrt{\frac{16}{9}} = \frac{4}{3}\)
     • Thus, \(\sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}\) confirms \(x = \frac{9}{25}\) is correct.
   • Similarly, for \(x = \frac{16}{25}\):
     • \(\sqrt{\frac{x}{1-x}} = \sqrt{\frac{\frac{16}{25}}{\frac{9}{25}}} = \frac{4}{3}\)
     • \(\sqrt{\frac{1-x}{x}} = \sqrt{\frac{\frac{9}{25}}{\frac{16}{25}}} = \frac{3}{4}\)
     • This also gives \(\frac{25}{12}\) as the sum, confirming \(x = \frac{16}{25}\) is correct.

Thus, the correct values of \(x\) that satisfy the equation are \(\frac{9}{25}\) or \(\frac{16}{25}\).