6. Problem: For what values of the parameter $a$ does the equation have a unique solution $\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}+x=a ?$
Solution: Notice that the expression under the square root can be represented as a complete square.
$x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=x+\frac{1}{4}+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+\frac{1}{4}=\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}$, then the equation will take the form $\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}}+x=a$, extracting the square root

$\left|\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right|+x=a$, the expression under the modulus is always greater than zero, therefore

$\sqrt{x+\frac{1}{4}}+\frac{1}{2}+x=a$

The left side of the equation again represents a complete square.

$\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}=a$. The equation has solutions only when $a \geq 0$.

1) If $a=0$, then $\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}=0, \quad \sqrt{x+\frac{1}{4}}+\frac{1}{2}=0, \quad \sqrt{x+\frac{1}{4}}=-\frac{1}{2}$ and there are no solutions.
2) If $a>0$, then $\sqrt{x+\frac{1}{4}}+\frac{1}{2}= \pm \sqrt{a}, \sqrt{x+\frac{1}{4}}=-\frac{1}{2} \pm \sqrt{a}$.

The equation $\sqrt{x+\frac{1}{4}}=-\frac{1}{2}-\sqrt{a}$ has no solutions, as the right side is less than zero.

The equation $\sqrt{x+\frac{1}{4}}=-\frac{1}{2}+\sqrt{a}$ has a solution if $-\frac{1}{2}+\sqrt{a} \geq 0$, that is, $a \geq \frac{1}{4}$. $x+\frac{1}{4}=\left(-\frac{1}{2}+\sqrt{a}\right)^{2}, \quad x=-\frac{1}{4}+\left(-\frac{1}{2}+\sqrt{a}\right)^{2}$ for the specified values of $a$ the solution is unique.

Answer: $a \in\left[\frac{1}{4} ;+\infty\right)$