(3) \( \sqrt{4 x+1}=3+\sqrt{4 x-2} \)

To solve the equation \( \sqrt{4 x + 1} = 3 + \sqrt{4 x - 2} \), start by squaring both sides to eliminate the square roots. This leads to: \[ 4x + 1 = (3 + \sqrt{4x - 2})^2 \] Expanding the right side results in: \[ 4x + 1 = 9 + 6\sqrt{4x - 2} + (4x - 2) \] Combine like terms: \[ 4x + 1 = 4x + 7 + 6\sqrt{4x - 2} \] Subtracting \(4x + 7\) from both sides gives: \[ 1 - 7 = 6\sqrt{4x - 2} \] Which simplifies to: \[ -6 = 6\sqrt{4x - 2} \] Dividing both sides by 6: \[ -1 = \sqrt{4x - 2} \] Since the square root cannot be negative, this means there are no real solutions to the equation. Therefore, the final conclusion is: \[ \text{No real solution} \]