If $$x$$ is a solution of the equation $$\sqrt{2x+1} - \sqrt{2x-1} = 1$$, $$\left(x \geq \frac{1}{2}\right)$$, then $$\sqrt{4x^2 - 1}$$ is equal to:

We begin with the given equation

$$\sqrt{2x+1}\;-\;\sqrt{2x-1}\;=\;1,\qquad x\;\ge\;\dfrac12.$$

To remove the surds, we first isolate the square roots on one side and then square both sides. The algebraic identity that we shall use is

$$(A-B)^2 \;=\; A^2 + B^2 - 2AB.$$

Putting $$A=\sqrt{2x+1}$$ and $$B=\sqrt{2x-1},$$ we have

$$\left(\sqrt{2x+1}-\sqrt{2x-1}\right)^2 \;=\; 1^2.$$

Expanding the left‐hand side with the identity, we get

$$\bigl(2x+1\bigr) + \bigl(2x-1\bigr) \;-\; 2\sqrt{(2x+1)(2x-1)} \;=\; 1.$$

Adding the first two terms gives $$4x,$$ so

$$4x \;-\; 2\sqrt{(2x+1)(2x-1)} \;=\; 1.$$

Let us introduce a convenient symbol for the remaining radical. Write

$$\sqrt{(2x+1)(2x-1)} \;=\;\sqrt{4x^2-1}.$$

Denote this expression by $$S,$$ i.e.

$$S \;=\;\sqrt{4x^2-1}.$$

With this notation the previous equation becomes

$$4x \;-\; 2S \;=\; 1.$$

Now we can solve for $$S$$ directly:

$$2S \;=\; 4x \;-\; 1 \;\;\Longrightarrow\;\; S \;=\; 2x \;-\;\dfrac12.$$

Remember that $$S=\sqrt{4x^2-1},$$ so we have obtained the relation

$$\sqrt{4x^2-1} \;=\; 2x \;-\; \dfrac12.$$

To find the numerical value, we must determine $$x.$$ Therefore we square both sides again. First write

$$4x^2 - 1 \;=\; \left(2x - \dfrac12\right)^2.$$

Expanding the right‐hand side gives

$$\left(2x - \dfrac12\right)^2 \;=\; (2x)^2 \;-\; 2\cdot 2x \cdot \dfrac12 \;+\; \left(\dfrac12\right)^2 \;=\; 4x^2 \;-\; 2x \;+\; \dfrac14.$$

Thus we have the equality

$$4x^2 - 1 \;=\; 4x^2 \;-\; 2x \;+\; \dfrac14.$$

Subtract $$4x^2$$ from both sides to eliminate the common term:

$$-1 \;=\; -2x \;+\; \dfrac14.$$

Now move all constants to one side:

$$-2x \;=\; -1 \;-\; \dfrac14 \;=\; -\dfrac54.$$

Dividing by $$-2$$ gives

$$x \;=\; \dfrac54 \times \dfrac12 \;=\; \dfrac58.$$

We check the domain condition $$x\ge \dfrac12;$$ clearly $$\dfrac58>\dfrac12,$$ so the value is admissible.

Finally, we evaluate the required quantity:

$$\sqrt{4x^2-1} \;=\;\sqrt{4\left(\dfrac58\right)^2 - 1} \;=\;\sqrt{4\cdot \dfrac{25}{64} - 1} \;=\;\sqrt{\dfrac{100}{64} - \dfrac{64}{64}} \;=\;\sqrt{\dfrac{36}{64}} \;=\;\sqrt{\left(\dfrac34\right)^2} \;=\;\dfrac34.$$

Hence, the correct answer is Option A.